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      <video:title>Problem 16-1: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>The angular velocity of a disk is defined by $\omega = (5t^2 + 2)$ rad/s, where $t$ is in seconds. The radius of the disk is 0.8 m. Determine the magnitudes of the velocity and acceleration of point A on the rim of the disk when $t = 0.5$ s.</video:description>
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    <lastmod>2026-07-25</lastmod>
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      <video:title>Problem 16-2: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>The angular acceleration of a disk is defined by $\alpha = 3t^2 + 12$ rad/s², where $t$ is in seconds. The disk is originally rotating at $\omega_0 = 12$ rad/s. Determine the magnitude of the velocity and the normal and tangential components of acceleration of point A on the disk when $t = 2$ s. Point A is located $r = 0.5$ m from the center of the disk.</video:description>
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    <lastmod>2026-07-25</lastmod>
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      <video:title>Problem 16-3: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>A disk originally rotating at $\omega_0 = 12 \ \text{rad/s}$ is subjected to a constant angular acceleration of $\alpha = 20 \ \text{rad/s}^2$. Point A is located at a radial distance of $r = 0.5 \ \text{m}$ from the center of the disk. Determine the magnitudes of the velocity and the normal and tangential components of acceleration of point A at the instant $t = 2 \ \text{s}$.</video:description>
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    <loc>https://doassignment.ca/videos/b1-1-explore-the-development-and-use-of-a-number-concept-concept-lesson-xstkolocack</loc>
    <lastmod>2026-09-05</lastmod>
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      <video:title>B1.1 — Explore the development and use of a number concept | Concept Lesson</video:title>
      <video:description>Question: Explore the concept of zero and its properties: the additive identity, multiplication by zero, and division involving zero. Then evaluate the following expression by applying each property: E equals the quantity 5 plus 0, times the quantity 3 minus 3, all divided by 2, plus 7 times 0, minus the quantity negative 4 plus 0. B1.1 — Explore the development and use of a number concept | Concept Lesson This lesson explores zero as a number concept, covering its historical development and three core properties: the additive identity, multiplication by zero, and division involving zero. Students work through a multi-step expression that applies all three rules and examine the critical distinction between dividing zero by a number versus dividing a number by zero. In this lesson, you will learn to: • Explain the additive identity property of zero and apply it to simplify expressions</video:description>
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      <video:publication_date>2026-09-05T20:15:01.000Z</video:publication_date>
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    <loc>https://doassignment.ca/videos/a1-apply-mathematical-processes-and-communicate-reasoning-concept-lesson-ipkdrp6li-0</loc>
    <lastmod>2026-09-05</lastmod>
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      <video:title>A1 — Apply mathematical processes and communicate reasoning | Concept Lesson</video:title>
      <video:description>Question: A cell-phone plan charges a flat monthly fee of 18 dollars plus 5 cents per text message sent. A second plan charges 11 cents per text message with no flat fee. How many text messages per month make the two plans cost the same? Justify your answer fully using the four-part framework. A1 — Apply mathematical processes and communicate reasoning | Concept Lesson This lesson introduces the four-part justification framework — State, Set Up, Solve, Verify and Conclude — as a structured approach to applying and communicating mathematical reasoning. Students work through a linear-equation break-even problem and learn to interpret their answer in context. The lesson directly supports Ontario Grade 9 curriculum expectation A1: Apply Mathematical Processes and Communicate Reasoning. In this lesson, you will learn to: • Apply the four-part justification framework to present a complete,</video:description>
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      <video:publication_date>2026-09-05T20:15:30.000Z</video:publication_date>
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    <loc>https://doassignment.ca/videos/a1-apply-mathematical-processes-and-communicate-reasoning-worked-example-taappbcwm-u</loc>
    <lastmod>2026-09-05</lastmod>
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      <video:title>A1 — Apply mathematical processes and communicate reasoning | Worked Example</video:title>
      <video:description>Question: Two straight paths run through a park. Path A passes through the points negative 3 comma 1 and 3 comma 5. Path B passes through the point 0 comma negative 4 and has a slope three times the slope of Path A. Part a: Write the equation of each path in slope-intercept form. Part b: Find the coordinates of the point where the two paths intersect. Part c: Determine which path is steeper and explain your reasoning using the concept of slope. A1 — Apply mathematical processes and communicate reasoning | Worked Example This lesson covers writing equations of lines in slope-intercept form, solving a system of two linear equations to find their intersection point, and comparing the steepness of lines using the concept of slope. Students work through a real-world context involving two straight paths on a coordinate plane. In this lesson, you will learn to: • Calculate the slope of a lin</video:description>
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      <video:publication_date>2026-09-05T20:15:25.000Z</video:publication_date>
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    <loc>https://doassignment.ca/videos/a2-connect-mathematics-to-real-life-knowledge-systems-and-careers-concept-lesson-lhuso2tvpvs</loc>
    <lastmod>2026-09-05</lastmod>
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      <video:title>A2 — Connect mathematics to real life, knowledge systems, and careers | Concept Lesson</video:title>
      <video:description>Question: A farm water tank starts the week with 2400 litres. Livestock consume a steady 350 litres per day. After how many complete days will the water level fall below 500 litres, signalling that the tank must be refilled? A2 — Connect mathematics to real life, knowledge systems, and careers | Concept Lesson This lesson explores Ontario Grade 9 curriculum expectation A2 by grounding the linear equation y = mx + b in a real-world resource-management scenario. Students model a decreasing water supply, solve a linear inequality to find a critical threshold day, and connect the mathematics to careers and diverse knowledge traditions. In this lesson, you will learn to: • Identify the slope and initial value in a real-world linear relationship and write the corresponding equation in the form y = mx + b. • Solve a linear inequality involving a negative rate of change, correctly reversing t</video:description>
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      <video:publication_date>2026-09-05T20:15:19.000Z</video:publication_date>
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    <loc>https://doassignment.ca/videos/a2-connect-mathematics-to-real-life-knowledge-systems-and-careers-worked-example-uhgqcm49fee</loc>
    <lastmod>2026-09-05</lastmod>
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      <video:title>A2 — Connect mathematics to real life, knowledge systems, and careers | Worked Example</video:title>
      <video:description>Question: A community garden charges a one-time registration fee of twelve dollars and then seven dollars per month for a garden plot. A neighbouring allotment program charges no registration fee but costs ten dollars per month. Part a: Write a linear equation for the total cost C, in dollars, of each program after m months. Part b: Graph both cost functions on the same grid. Part c: Find the number of months at which both programs cost the same. Part d: A gardener plans to use a plot for exactly five months. Which program is cheaper, and by how much? A2 — Connect mathematics to real life, knowledge systems, and careers | Worked Example This lesson uses two real-world pricing plans to build and compare linear equations in slope-intercept form. Students graph both cost functions, solve algebraically for the break-even point, and interpret the solution to make a financially informed deci</video:description>
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      <video:publication_date>2026-09-05T20:15:05.000Z</video:publication_date>
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    <lastmod>2026-08-06</lastmod>
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      <video:title>Finding the Intersection Point by Graphing Two Lines</video:title>
      <video:description>Question: Graph the system of equations and find the solution. Identify the point of intersection and verify by substituting into both equations. The first equation is y equals 2x minus 3. The second equation is y equals negative x plus 6. Finding the Intersection Point by Graphing Two Lines This lesson teaches students how to solve a system of two linear equations by graphing both lines on the same coordinate plane and identifying their point of intersection. Students also learn to confirm the graphical solution algebraically and verify it by substitution into both original equations. In this lesson, you will learn to: • Identify the slope and y-intercept of a linear equation written in slope-intercept form • Graph two linear equations on the same coordinate plane and identify their intersection point • Solve a system of linear equations algebraically by setting the expressions equal</video:description>
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      <video:publication_date>2026-08-06T22:25:37.000Z</video:publication_date>
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      <video:live>no</video:live>
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    <loc>https://doassignment.ca/videos/identifying-systems-with-no-solution-or-infinitely-many-solutions-rj5xpeg3jpk</loc>
    <lastmod>2026-08-06</lastmod>
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      <video:title>Identifying Systems with No Solution or Infinitely Many Solutions</video:title>
      <video:description>Question: Graph the following system of equations and determine the number of solutions. Explain whether the system has no solution, one solution, or infinitely many solutions. Equation 1: 3x minus 2y equals 8. Equation 2: 6x minus 4y equals 20. Identifying Systems with No Solution or Infinitely Many Solutions This lesson explores how to determine the number of solutions in a linear system by rewriting equations in slope-intercept form and comparing slopes and y-intercepts. Students learn that parallel lines — same slope, different intercepts — produce an inconsistent system with no solution, confirmed both graphically and algebraically. In this lesson, you will learn to: • Rewrite linear equations in slope-intercept form by solving for y • Identify parallel lines by comparing slopes and y-intercepts • Determine that a system with parallel lines has no solution (inconsistent system) •</video:description>
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      <video:publication_date>2026-08-06T22:25:37.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
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  <url>
    <loc>https://doassignment.ca/videos/real-world-application-break-even-point-analysis-to2ry-ubppa</loc>
    <lastmod>2026-08-06</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/to2RY-UbPPA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Real-World Application: Break-Even Point Analysis</video:title>
      <video:description>Question: A small bakery has a cost equation C equals 150 plus 4x, and a revenue equation R equals 9x, where x is the number of loaves of bread. Graph both equations on the same coordinate plane and find the break-even point where cost equals revenue. Interpret what this point means in context. Real-World Application: Break-Even Point Analysis This lesson uses a bakery cost-and-revenue scenario to explore systems of linear equations. Students graph both equations on the same coordinate plane, solve the system algebraically by setting the two expressions equal, and interpret the intersection point as the break-even point in context. In this lesson, you will learn to: • Identify and write a system of two linear equations from a real-world context • Solve a system of linear equations algebraically using the substitution/elimination method • Graph two linear equations on the same coordina</video:description>
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      <video:publication_date>2026-08-06T22:25:37.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
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  <url>
    <loc>https://doassignment.ca/videos/finding-slope-intercept-form-from-two-points-esxi96bh99w</loc>
    <lastmod>2026-07-31</lastmod>
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      <video:thumbnail_loc>https://i.ytimg.com/vi/Esxi96Bh99w/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Finding Slope-Intercept Form from Two Points</video:title>
      <video:description>Question: A line passes through the points 2 comma 5 and 6 comma 17. Write the equation of the line in slope-intercept form, y equals m x plus b. Show all steps to find the slope and y-intercept. Finding Slope-Intercept Form from Two Points This lesson walks through finding the equation of a line in slope-intercept form given two points. Students learn to calculate slope using the slope formula, then solve for the y-intercept by substituting a known point, and finally verify the equation with a second point and a graph. In this lesson, you will learn to: • Calculate the slope of a line given two coordinate points using the slope formula. • Determine the y-intercept by substituting a known point into the slope-intercept form. • Write and verify the equation of a line in slope-intercept form y = mx + b. • Interpret the slope and y-intercept graphically on a coordinate plane. Course: Al</video:description>
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      <video:publication_date>2026-07-31T19:08:42.000Z</video:publication_date>
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    <loc>https://doassignment.ca/videos/graphing-and-interpreting-slope-intercept-form-ecpekfoxh4k</loc>
    <lastmod>2026-07-31</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/ECPeKfOXh4k/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Graphing and Interpreting Slope-Intercept Form</video:title>
      <video:description>Question: The equation y equals negative 2 x plus 10 represents a linear function. Part a: Identify the slope and y-intercept. Part b: Graph the line on a coordinate plane, clearly marking the y-intercept and at least two other points using the slope. Part c: Find the x-intercept by setting y equal to 0 and solving for x. Part d: Describe what the slope tells us about the graph&apos;s direction and steepness. Graphing and Interpreting Slope-Intercept Form This lesson explores the linear equation y = −2x + 10 as a model for understanding slope-intercept form. Students learn to identify the slope and y-intercept, build a table of points, graph the line, find the x-intercept algebraically, and interpret what slope communicates about a line&apos;s direction and steepness. In this lesson, you will learn to: • Identify the slope and y-intercept from an equation written in slope-intercept form y = mx</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/ECPeKfOXh4k</video:player_loc>
      <video:publication_date>2026-07-31T19:08:44.000Z</video:publication_date>
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    <loc>https://doassignment.ca/videos/writing-equations-from-real-world-scenarios-fhqbmmvpeck</loc>
    <lastmod>2026-07-31</lastmod>
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      <video:title>Writing Equations from Real-World Scenarios</video:title>
      <video:description>Question: A swimming pool is being drained. It starts with 8,000 gallons of water and drains at a constant rate of 250 gallons per hour. Write an equation in slope-intercept form that represents the amount of water, y in gallons, remaining in the pool after t hours. Then determine how much water remains after 12 hours. Writing Equations from Real-World Scenarios This lesson uses a real-world draining scenario to build a linear equation in slope-intercept form, y = mt + b. Students identify the initial value as the y-intercept and a constant rate of change as the slope, then evaluate the equation at a specific input value and verify the result. In this lesson, you will learn to: • Identify the slope and y-intercept from a real-world context and write an equation in slope-intercept form • Interpret the meaning of a negative slope as a constant decrease in a practical situation • Evaluat</video:description>
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      <video:publication_date>2026-07-31T19:08:44.000Z</video:publication_date>
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  <url>
    <loc>https://doassignment.ca/videos/finding-linear-equation-from-point-and-slope-8brbmz89gcg</loc>
    <lastmod>2026-07-31</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/8BrBmZ89GCg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Finding Linear Equation from Point and Slope</video:title>
      <video:description>Question: A line passes through the point 3 comma 7 and has a slope of 2. Write the equation of the line in point-slope form, then convert it to slope-intercept form. Finding Linear Equation from Point and Slope This lesson shows how to write the equation of a line given its slope and one point on it. Students learn to apply the point-slope formula first, then convert to slope-intercept form through distribution and solving for y. In this lesson, you will learn to: • Apply the point-slope formula using a given slope and point • Convert a point-slope equation to slope-intercept form by distributing and isolating y • Identify the slope and y-intercept from slope-intercept form • Verify a linear equation by substituting the given point Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 4- Writing Linear Functions Level: Grade 9–10 Pause before each step, try the next part yoursel</video:description>
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      <video:publication_date>2026-07-31T19:50:32.000Z</video:publication_date>
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    <loc>https://doassignment.ca/videos/writing-equation-from-two-points-wj6g7cj5xwi</loc>
    <lastmod>2026-07-31</lastmod>
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      <video:thumbnail_loc>https://i.ytimg.com/vi/WJ6g7cJ5xWI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Writing Equation from Two Points</video:title>
      <video:description>Question: Two points on a line are negative 2 comma 5 and 4 comma negative 1. Find the slope of the line, write the equation in point-slope form using one of the points, and then express it in slope-intercept form. Writing Equation from Two Points This lesson walks through finding the slope from two coordinate points, then expressing the line in both point-slope form and slope-intercept form. Students practice the full workflow from the slope formula through algebraic manipulation to graphing. In this lesson, you will learn to: • Calculate the slope of a line given two points using the slope formula. • Write the equation of a line in point-slope form using a known point and slope. • Convert a point-slope equation to slope-intercept form by solving for y. • Graph a linear equation and verify that given points lie on the line. Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 4-</video:description>
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      <video:publication_date>2026-07-31T19:50:36.000Z</video:publication_date>
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    <loc>https://doassignment.ca/videos/finding-parallel-line-equations-from-a-given-slope-5eyenndbn00</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
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      <video:title>Finding Parallel Line Equations from a Given Slope</video:title>
      <video:description>Question: A line passes through the point 2 comma 5 and is parallel to the line y equals 3x minus 4. Write the equation of the line in slope-intercept form. Finding Parallel Line Equations from a Given Slope This lesson covers how to write the equation of a line that is parallel to a given line and passes through a specific point. Students use the parallel lines slope rule and the slope-intercept form to determine the unknown y-intercept by substituting the given point. In this lesson, you will learn to: • Identify the slope of a line written in slope-intercept form. • Apply the parallel lines rule that parallel lines have equal slopes. • Use a known point to solve for the y-intercept in slope-intercept form. • Write and verify the equation of a line in slope-intercept form. Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 4- Writing Linear Functions Level: Grade 9 Pause bef</video:description>
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      <video:publication_date>2026-08-03T10:15:40.000Z</video:publication_date>
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    <loc>https://doassignment.ca/videos/finding-perpendicular-line-equations-using-negative-reciprocals-jv3f2v040ak</loc>
    <lastmod>2026-08-03</lastmod>
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      <video:title>Finding Perpendicular Line Equations Using Negative Reciprocals</video:title>
      <video:description>Question: Write the equation of the line that passes through the point negative 3 comma 2 and is perpendicular to y equals 2x plus 6. Express your answer in slope-intercept form. Finding Perpendicular Line Equations Using Negative Reciprocals This lesson covers how to write the equation of a line perpendicular to a given line through a specific point. Students learn to find the negative reciprocal slope, apply point-slope form, and convert to slope-intercept form. In this lesson, you will learn to: • Identify the slope of a line written in slope-intercept form • Calculate the negative reciprocal slope for a perpendicular line • Apply point-slope form to write a line through a given point • Verify perpendicularity by confirming the product of the slopes equals −1 Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 4- Writing Linear Functions Level: Grade 9 Pause before each step</video:description>
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      <video:publication_date>2026-08-03T10:15:35.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/identifying-parallel-and-perpendicular-lines-from-equations-rad3gcu-vi</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/_rAD3GCu-VI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Identifying Parallel and Perpendicular Lines from Equations</video:title>
      <video:description>Question: Determine whether the lines 4x minus 2y equals 8, y equals 2x plus 3, and 3y equals negative 6x plus 9 are parallel, perpendicular, or neither to each other. Justify your answer by comparing slopes. Identifying Parallel and Perpendicular Lines from Equations This lesson teaches students how to determine whether pairs of lines are parallel, perpendicular, or neither by converting each equation into slope-intercept form and comparing slopes. Students practice rewriting standard-form and non-standard equations, then apply the rules that parallel lines share equal slopes and perpendicular lines have slopes whose product is −1. In this lesson, you will learn to: • Convert linear equations from standard form into slope-intercept form y = mx + b. • Identify the slope and y-intercept of a line from its slope-intercept equation. • Apply the parallel-slope rule (equal slopes, differen</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/_rAD3GCu-VI</video:player_loc>
      <video:publication_date>2026-08-03T10:15:30.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/finding-the-line-of-best-fit-from-temperature-data-l-k6vdpupf0</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/L-k6VDpUpf0/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Finding the Line of Best Fit from Temperature Data</video:title>
      <video:description>Question: A weather station records the daily high temperature in degrees Fahrenheit and the number of ice cream cones sold at a nearby shop over 6 days. The data points are: 68 comma 42, 72 comma 58, 75 comma 71, 78 comma 85, 82 comma 103, and 85 comma 118. Create a scatter plot of the data, identify the correlation, and find the equation of the line of best fit using two reasonable points from the data. Then predict how many ice cream cones would be sold if the high temperature reaches 90 degrees Fahrenheit. Finding the Line of Best Fit from Temperature Data This lesson explores how to create a scatter plot from a two-variable data set, identify the type and strength of correlation, and write the equation of the line of best fit using two representative points. Students then use the equation to make predictions beyond the recorded data. In this lesson, you will learn to: • Plot two-</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/L-k6VDpUpf0</video:player_loc>
      <video:publication_date>2026-08-03T18:59:18.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/writing-a-linear-function-from-a-real-world-scenario-91tkzbwejd8</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/91tKzbwEJd8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Writing a Linear Function from a Real-World Scenario</video:title>
      <video:description>Question: A gym membership costs 25 dollars per month after an initial enrollment fee. A member pays a total of 160 dollars after 4 months and 285 dollars after 10 months. Part a: Write a linear equation in the form y equals m x plus b, where y represents total cost and x represents number of months. Part b: Identify the enrollment fee and monthly rate. Part c: How much will a member pay in total after 18 months? Writing a Linear Function from a Real-World Scenario This lesson guides students through building a linear equation in slope-intercept form from two given data points representing a gym membership scenario. Students learn to interpret the slope as a rate of change and the y-intercept as a fixed starting cost, then use the equation to make predictions. In this lesson, you will learn to: • Write a linear equation in slope-intercept form using two given data points • Interpret t</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/91tKzbwEJd8</video:player_loc>
      <video:publication_date>2026-08-03T18:59:14.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/interpreting-and-using-linear-equations-in-context-pc4jif-4ima</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/PC4jiF-4iMA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Interpreting and Using Linear Equations in Context</video:title>
      <video:description>Question: A phone plan charges a base rate plus a per-minute charge. The total monthly bill is modeled by the equation C equals 12 plus 0.08 times m, where C is the cost in dollars and m is the number of minutes used. Part a: Identify and interpret the y-intercept in context. Part b: Identify and interpret the slope in context. Part c: If the monthly bill is 28 dollars and 80 cents, how many minutes were used? Part d: Create a table showing costs for 0, 50, 100, 150, and 200 minutes, then verify your answer to part c using the table or graph. Interpreting and Using Linear Equations in Context This lesson explores a real-world linear equation in slope-intercept form by identifying and interpreting the slope and y-intercept in context. Students solve for an unknown input given an output value, build a table of values, and verify their answer using a graph. In this lesson, you will learn</video:description>
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      <video:publication_date>2026-08-03T18:59:10.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/finding-the-best-fit-line-for-temperature-data-qhya9d-u6m4</loc>
    <lastmod>2026-08-04</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/QHYA9D_u6m4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Finding the Best-Fit Line for Temperature Data</video:title>
      <video:description>Question: A data set records the average daily temperature in degrees Fahrenheit and the number of ice cream sales for 6 consecutive days. Day 1: 62 degrees, 18 sales. Day 2: 68 degrees, 35 sales. Day 3: 75 degrees, 52 sales. Day 4: 71 degrees, 45 sales. Day 5: 79 degrees, 68 sales. Day 6: 85 degrees, 81 sales. Part a: Create a scatter plot with temperature on the x-axis and sales on the y-axis. Part b: Draw a line of best fit and estimate its equation in slope-intercept form. Part c: Use your equation to predict ice cream sales when the temperature is 90 degrees Fahrenheit. Part d: Explain whether your prediction is reliable and why. Finding the Best-Fit Line for Temperature Data This lesson teaches students how to represent bivariate data on a scatter plot, draw and write the equation of a line of best fit in slope-intercept form, and use the equation to make predictions. Students al</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/QHYA9D_u6m4</video:player_loc>
      <video:publication_date>2026-08-04T18:58:51.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/comparing-two-lines-of-fit-for-study-hours-vs-test-scores-o7yqf8w94q0</loc>
    <lastmod>2026-08-04</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/o7yQF8W94Q0/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Comparing Two Lines of Fit for Study Hours vs. Test Scores</video:title>
      <video:description>Question: A data set records the number of hours studied and the corresponding test scores. Study hours go from 1 to 8. The test scores for those hours are 52, 61, 58, 78, 82, 87, 89, and 95. Part a: Plot the data and identify any points that appear to be outliers. Part b: Determine a line of best fit for all 8 data points, and write the equation in the form y equals m x plus b. Part c: Determine a second line of best fit by removing the outlier or outliers, and write that equation. Part d: Compare the two lines. Which equation would you recommend for making predictions? Justify your choice with at least two reasons. Comparing Two Lines of Fit for Study Hours vs. Test Scores This lesson explores how to find a least-squares line of best fit for a scatter plot, identify outliers visually and numerically, and compare two regression lines — one with and one without the outlier — to determi</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/o7yQF8W94Q0</video:player_loc>
      <video:publication_date>2026-08-04T18:58:48.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/profit-vs-production-volume-line-of-fit-0f68a-vglrm</loc>
    <lastmod>2026-08-04</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/0f68A-VglrM/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Profit vs. Production Volume Line of Fit</video:title>
      <video:description>Question: A manufacturing company tracks its monthly profit in thousands of dollars based on the number of units produced in hundreds. The data is as follows: when 10 hundred units are produced, the monthly profit is 5 thousand dollars; 15 hundred units gives 12 thousand dollars profit; 20 hundred units gives 18 thousand dollars; 25 hundred units gives 26 thousand dollars; 30 hundred units gives 32 thousand dollars; and 35 hundred units gives 40 thousand dollars. Part a: Plot the data points on a coordinate grid. Part b: Use two points on or near your trend line to find the slope. Part c: Write the equation of the line of best fit. Part d: Interpret the slope in context — what does it tell you about the company&apos;s profit per 100 units produced? Part e: If the company produces 4000 units, predict the monthly profit. Profit vs. Production Volume Line of Fit This lesson guides students thr</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/0f68A-VglrM</video:player_loc>
      <video:publication_date>2026-08-04T18:58:43.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/finding-the-common-difference-and-nth-term-dm01sdsu2xc</loc>
    <lastmod>2026-08-05</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/dm01Sdsu2xc/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Finding the Common Difference and nth Term</video:title>
      <video:description>Question: A swimming pool&apos;s water level is being monitored daily. On day 1, the level is 48 inches. On day 2, it is 46 inches. On day 3, it is 44 inches. Identify the common difference and write an explicit formula for the water level after n days. Then find the water level on day 12. Finding the Common Difference and nth Term This lesson walks through identifying the common difference of an arithmetic sequence, building an explicit formula of the form $a_n = a_1 + (n - 1)d$, and using that formula to find any term. A real-world decreasing sequence is used to connect arithmetic sequences to linear functions. In this lesson, you will learn to: • Identify the common difference of an arithmetic sequence by subtracting consecutive terms. • Write an explicit formula for an arithmetic sequence using the formula $a_n = a_1 + (n - 1)d$. • Simplify and verify an explicit formula against known</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/dm01Sdsu2xc</video:player_loc>
      <video:publication_date>2026-08-05T11:37:23.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/writing-a-linear-function-from-an-arithmetic-sequence-gfnohseqcqs</loc>
    <lastmod>2026-08-05</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/GfNOHSeQcqs/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Writing a Linear Function from an Arithmetic Sequence</video:title>
      <video:description>Question: A video streaming service charges a one-time setup fee plus a monthly subscription. In month 1, the total cost is 25 dollars. In month 3, the total cost is 55 dollars. In month 6, the total cost is 100 dollars. Determine the monthly fee, which is the slope, the one-time setup cost, which is the y-intercept, and write a linear function C of m for the total cost after m months. What is the total cost after one year? Writing a Linear Function from an Arithmetic Sequence This lesson uses a streaming service pricing scenario to build a linear function from given data points. Students identify slope and y-intercept from a table of values, write the function in slope-intercept form, and use it to make predictions. In this lesson, you will learn to: • Calculate the slope (rate of change) from two data points • Determine the y-intercept by substituting a known point into the linear e</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/GfNOHSeQcqs</video:player_loc>
      <video:publication_date>2026-08-05T11:37:26.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/recursive-and-explicit-formulas-in-real-world-context-f467pdvwqys</loc>
    <lastmod>2026-08-05</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/F467pdVwQys/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Recursive and Explicit Formulas in Real-World Context</video:title>
      <video:description>Question: A concert venue adds seats in arithmetic sequence rows. Row 8 has 72 seats and row 12 has 88 seats. Write both a recursive formula — a of n equals a of n minus 1, plus d — and an explicit formula — a of n equals a of 1, plus the quantity n minus 1 times d — for the number of seats in row n. Then find the total number of seats in the first 15 rows. Recursive and Explicit Formulas in Real-World Context This lesson connects arithmetic sequences to linear functions by finding recursive and explicit formulas from two known terms. Students apply the arithmetic series sum formula to find a total, reinforcing the relationship between sequences, slope, and linear equations. In this lesson, you will learn to: • Find the common difference of an arithmetic sequence from two non-consecutive terms • Write both recursive and explicit formulas for an arithmetic sequence • Verify a formula u</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/F467pdVwQys</video:player_loc>
      <video:publication_date>2026-08-05T11:37:30.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/calculus-quadratic-equations-practice-set-1-1wd-nw2vvd4</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/1WD-NW2vvD4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Calculus – Quadratic Equations Practice Set · 1</video:title>
      <video:description>Question: Which of the following are the solutions to x squared minus 5x plus 6 equals zero? Calculus – Quadratic Equations Practice Set · 1 This lesson demonstrates how to solve a quadratic equation of the form $x^2 + bx + c = 0$ by factoring it into two binomials. Students learn to identify a factor pair, apply the Zero Product Property, and verify both solutions by substitution. In this lesson, you will learn to: • Find a factor pair of integers that satisfies both a sum and product condition to factor a quadratic trinomial. • Apply the Zero Product Property to set each binomial factor equal to zero and solve for the variable. • Verify solutions by substituting each value back into the original equation. Course: Calculus Level: Grade 9 Pause before each step, try the next part yourself, and then continue to check your reasoning. Find more learning support: https://doassignment.c</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/1WD-NW2vvD4</video:player_loc>
      <video:publication_date>2026-08-03T19:06:09.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/calculus-quadratic-equations-practice-set-2-shvla-fehju</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/ShVLA-FEHJU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Calculus – Quadratic Equations Practice Set · 2</video:title>
      <video:description>Question: Use the quadratic formula to solve 2 x squared minus 4 x minus 6 equals 0. Find both solutions. Calculus – Quadratic Equations Practice Set · 2 This lesson demonstrates how to apply the quadratic formula to solve a quadratic equation with integer coefficients. Students learn to identify coefficients, compute the discriminant, and simplify both roots. Verification of each solution by substitution is also covered. In this lesson, you will learn to: • Identify the coefficients a, b, and c from a quadratic equation in standard form • Calculate the discriminant and use it to determine the nature and number of solutions • Apply the quadratic formula to find both roots of a quadratic equation • Verify solutions by substituting them back into the original equation Course: Calculus Level: Grade 9–10 Pause before each step, try the next part yourself, and then continue to check your</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/ShVLA-FEHJU</video:player_loc>
      <video:publication_date>2026-08-03T19:06:05.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/calculus-quadratic-equations-practice-set-3-ai6z-fbizig</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/ai6Z_fBIZig/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Calculus – Quadratic Equations Practice Set · 3</video:title>
      <video:description>Question: Find all values of x satisfying x squared plus 4x plus 13 equals 0, expressing each root in the form a plus b i, where a and b are real numbers. Enter both roots separated by a comma. Calculus – Quadratic Equations Practice Set · 3 This lesson solves a quadratic equation whose discriminant is negative, producing a conjugate pair of complex roots. Students learn to apply the quadratic formula, simplify square roots of negative numbers using the imaginary unit i, and express results in standard a + bi form. In this lesson, you will learn to: • Apply the quadratic formula to any quadratic equation and identify its coefficients • Compute and interpret a negative discriminant as an indicator of complex (non-real) roots • Simplify the square root of a negative number using the imaginary unit i • Express complex roots in standard a + bi form and verify them by expanding the factore</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/ai6Z_fBIZig</video:player_loc>
      <video:publication_date>2026-08-03T19:06:01.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/calculus-quadratic-equations-practice-set-4-hdeha71zbvk</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Hdeha71zbvk/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Calculus – Quadratic Equations Practice Set · 4</video:title>
      <video:description>Question: A particle moves along a straight line so that its position at time t greater than or equal to 0 is given by s of t equals negative t squared plus 8t minus 7, in metres. At what time t does the particle first reach a position of s equals 9 metres? Calculus – Quadratic Equations Practice Set · 4 This lesson explores how to find the time at which a particle reaches a specific position along a straight line, given a quadratic position function. Students set up and solve a quadratic equation, interpret a double root geometrically, and verify the answer by substitution. In this lesson, you will learn to: • Set up an equation by substituting a target position value into a quadratic position function • Rearrange and solve a quadratic equation by factoring a perfect-square trinomial • Interpret a double root as the parabola being tangent to a horizontal line • Verify a solution by s</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Hdeha71zbvk</video:player_loc>
      <video:publication_date>2026-08-03T19:05:51.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/calculus-linear-equations-practice-set-1-9ozq7rdfwq8</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/9OzQ7rdfWq8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Calculus – Linear Equations Practice Set · 1</video:title>
      <video:description>Question: Which of the following is the equation of the tangent line to f of x equals x squared plus 3x, at x equals 1? Calculus – Linear Equations Practice Set · 1 This lesson demonstrates how to find the equation of a tangent line to a polynomial function at a given point. Students apply the power rule to find the derivative, evaluate it for the slope, and use point-slope form to write the final equation. In this lesson, you will learn to: • Evaluate a function at a given x-value to find the point of tangency • Apply the power rule to differentiate a polynomial and obtain the slope function • Evaluate the derivative at a specific x-value to find the slope of the tangent line • Use point-slope form to write the equation of the tangent line Course: Calculus Level: Grade 12 Pause before each step, try the next part yourself, and then continue to check your reasoning. Find more learn</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/9OzQ7rdfWq8</video:player_loc>
      <video:publication_date>2026-08-03T19:05:10.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/calculus-linear-equations-practice-set-2-1ege5kqcwwe</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/1EgE5KqcWwE/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Calculus – Linear Equations Practice Set · 2</video:title>
      <video:description>Question: Find the slope of the tangent line to g of x equals the square root of x, at x equals 9. Calculus – Linear Equations Practice Set · 2 This lesson finds the slope of the tangent line to g(x) = √x at x = 9 using the limit definition of the derivative. The key technique is rationalizing the numerator with a conjugate to resolve the indeterminate form, then verifying the result with the Power Rule. In this lesson, you will learn to: • Apply the limit definition of the derivative to evaluate g′(a) for a given function and point • Rationalize a difference-of-square-roots expression in a difference quotient to eliminate an indeterminate form • Cancel common factors in the simplified difference quotient and evaluate the resulting limit by direct substitution • Verify a limit-definition result using the Power Rule Course: Calculus Level: Grade 12 Pause before each step, try the nex</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/1EgE5KqcWwE</video:player_loc>
      <video:publication_date>2026-08-03T19:05:05.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/calculus-linear-equations-practice-set-3-q-hdt8wsrm8</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/q_hdT8WSrm8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Calculus – Linear Equations Practice Set · 3</video:title>
      <video:description>Question: The position of a particle is given by s of t equals t cubed minus 6t squared plus 9t plus 2. Find the equation of the tangent line to s of t at t equals 2. Calculus – Linear Equations Practice Set · 3 This lesson finds the equation of the tangent line to a cubic position function at a specific point. Students apply the power rule to differentiate, evaluate the derivative for the slope, and use point-slope form to write the tangent line equation. In this lesson, you will learn to: • Differentiate a cubic polynomial using the power rule • Evaluate a function and its derivative at a given input value • Apply point-slope form to write the equation of a tangent line • Verify a tangent line equation by checking both the point and slope conditions Course: Calculus Level: Grade 12 Pause before each step, try the next part yourself, and then continue to check your reasoning. Find</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/q_hdT8WSrm8</video:player_loc>
      <video:publication_date>2026-08-03T19:04:50.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/calculus-linear-equations-practice-set-4-jntrf65chq8</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/JNtRf65chq8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Calculus – Linear Equations Practice Set · 4</video:title>
      <video:description>Question: Find the x-values at which the tangent line to h of x equals 2x cubed minus 12x is horizontal. Calculus – Linear Equations Practice Set · 4 This lesson shows how to find the x-values where a polynomial curve has a horizontal tangent line by setting the first derivative equal to zero and solving the resulting equation. The power rule is applied to differentiate a cubic function, and the solutions are verified by substitution. In this lesson, you will learn to: • Explain why a horizontal tangent line requires the derivative to equal zero • Apply the power rule to differentiate a cubic polynomial term by term • Solve a quadratic equation arising from setting the derivative to zero • Verify solutions by substituting back into the derivative Course: Calculus Level: Grade 12 Pause before each step, try the next part yourself, and then continue to check your reasoning. Find more</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/JNtRf65chq8</video:player_loc>
      <video:publication_date>2026-08-03T19:04:45.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/calculus-linear-equations-practice-set-5-kfszvmvidmo</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/kFsZVMviDMo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Calculus – Linear Equations Practice Set · 5</video:title>
      <video:description>Question: Let f of x equal e to the 2x. The tangent line to f at x equals 0 intersects the x-axis at which point? Calculus – Linear Equations Practice Set · 5 This lesson demonstrates how to find the equation of a tangent line to an exponential function at a given point using the chain rule. Students then locate where that tangent line crosses the x-axis by setting y equal to zero and solving. In this lesson, you will learn to: • Differentiate an exponential function using the chain rule • Write the equation of a tangent line using point-slope form • Find the x-intercept of a tangent line by setting y = 0 • Interpret the tangent line geometrically on a coordinate graph Course: Calculus Level: Grade 12 Pause before each step, try the next part yourself, and then continue to check your reasoning. Find more learning support: https://doassignment.ca This educational resource was creat</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/kFsZVMviDMo</video:player_loc>
      <video:publication_date>2026-08-03T19:04:40.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-quadratic-equations-with-a-perfect-square-trinomial-mc2xfdks7e8</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/MC2xFDKS7e8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Quadratic Equations with a Perfect Square Trinomial</video:title>
      <video:description>Question: Solve: x squared plus 2x plus 1 equals 0. Solving Quadratic Equations with a Perfect Square Trinomial This lesson shows how to solve a quadratic equation whose left side is a perfect square trinomial. Students learn to recognize the pattern, factor it into a squared binomial, and interpret the resulting double root. In this lesson, you will learn to: • Recognize a perfect square trinomial and apply the pattern $a^2 + 2ab + b^2 = (a+b)^2$ • Factor a quadratic equation into a squared binomial and solve by taking the square root • Understand what a repeated (double) root means in the context of a quadratic equation • Verify a solution by substituting it back into the original equation Course: Calculus Level: Grade 9 Pause before each step, try the next part yourself, and then continue to check your reasoning. Find more learning support: https://doassignment.ca This educatio</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/MC2xFDKS7e8</video:player_loc>
      <video:publication_date>2026-08-03T19:04:35.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/calculus-factoring-polynomials-practice-set-1-jttr0114oho</loc>
    <lastmod>2026-08-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/JTtR0114OHo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Calculus – Factoring Polynomials Practice Set · 1</video:title>
      <video:description>Question: Factor the polynomial p of x equals 6x cubed minus 9x squared minus 6x completely over the integers. Calculus – Factoring Polynomials Practice Set · 1 This lesson walks through the complete factorization of a cubic polynomial over the integers. Students first extract the greatest common factor, then factor the resulting quadratic trinomial using the AC method and grouping. The technique builds essential fluency for polynomial operations needed in precalculus and calculus. In this lesson, you will learn to: • Identify and factor out the greatest common factor (GCF) from a polynomial • Apply the AC method to factor a quadratic trinomial with a leading coefficient other than 1 • Use factor by grouping to complete the factorization of a quadratic • Write and verify the complete factored form of a cubic polynomial over the integers Course: Calculus Level: Grade 10 Pause before</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/JTtR0114OHo</video:player_loc>
      <video:publication_date>2026-08-03T19:04:31.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-a-sum-of-cubes-equation-jkjzfcbrrls</loc>
    <lastmod>2026-08-06</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/jKJZFCbrRls/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving a Sum of Cubes Equation</video:title>
      <video:description>Question: Solve for x: x cubed equals 30 cubed plus 40 cubed plus 50 cubed. Solving a Sum of Cubes Equation This lesson solves the equation x³ = 30³ + 40³ + 50³ by factoring out a common base and recognizing a perfect cube. Students discover the elegant identity 3³ + 4³ + 5³ = 6³ and apply properties of cube roots to find the solution. In this lesson, you will learn to: • Factor a common base from a sum of cubes to simplify expressions • Recognize and apply the identity 3³ + 4³ + 5³ = 6³ • Use the property (a · b)³ = a³ · b³ to combine factors • Solve an equation of the form x³ = k by taking cube roots Course: Calculus Level: Grade 9 Pause before each step, try the next part yourself, and then continue to check your reasoning. Find more learning support: https://doassignment.ca This educational resource was created from an anonymized session. It contains no student name, voice, fa</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/jKJZFCbrRls</video:player_loc>
      <video:publication_date>2026-08-06T21:58:06.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/making-an-input-output-table-graphing-a-linear-function-qyl0whv09ca</loc>
    <lastmod>2026-07-27</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/qyl0WHV09CA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Making an Input-Output Table &amp; Graphing a Linear Function</video:title>
      <video:description>Question: Complete the input-output table for the function y equals 2x minus 3 using the input values x equals negative 2, negative 1, 0, 1, and 2. Then plot the ordered pairs on a coordinate plane and describe whether the graph is linear. The table has three columns: x, y equals 2x minus 3, and the ordered pair x comma y. Fill in the output and ordered pair for each of the five input values. Making an Input-Output Table &amp; Graphing a Linear Function This lesson walks through evaluating a linear function by substituting input values into y = 2x − 3, recording the results in an input-output table, and plotting the ordered pairs on a coordinate plane. Students verify that all points lie on a straight line and connect the equation&apos;s slope and y-intercept to the graph&apos;s appearance. In this lesson, you will learn to: • Evaluate a linear function for given input values and complete an input-</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/qyl0WHV09CA</video:player_loc>
      <video:publication_date>2026-07-27T18:48:19.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/finding-the-slope-between-two-points-on-a-graph-gtv7heutwci</loc>
    <lastmod>2026-07-27</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/GTV7heUTWCI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Finding the Slope Between Two Points on a Graph</video:title>
      <video:description>Question: Find the slope of the line passing through each pair of points. Then describe the slope as positive, negative, zero, or undefined. Part a: the points 1 comma 3 and 4 comma 9. Part b: the points negative 2 comma 5 and 3 comma negative 5. Part c: the points negative 4 comma 2 and 6 comma 2. Part d: the points 3 comma negative 1 and 3 comma 7. Finding the Slope Between Two Points on a Graph This lesson covers how to calculate the slope of a line passing through two given points using the slope formula. Students practice identifying whether each slope is positive, negative, zero, or undefined by examining four representative examples. In this lesson, you will learn to: • Apply the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ to find the slope between two points • Distinguish between positive, negative, zero, and undefined slopes • Connect slope values to the visual behavior o</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/GTV7heUTWCI</video:player_loc>
      <video:publication_date>2026-07-27T18:48:22.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/real-world-linear-function-writing-interpreting-a-graph-hnoyqtf8eng</loc>
    <lastmod>2026-07-27</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/HNoYqTf8ENg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Real-World Linear Function: Writing &amp; Interpreting a Graph</video:title>
      <video:description>Question: A service charges a flat setup fee of 40 dollars plus 12 dollars per hour. Part a: Write a linear function that represents the total cost C in dollars as a function of hours h. Part b: Identify the slope and y-intercept and explain what each means in context. Part c: Create an input-output table for h equals 0, 1, 2, 3, 4, and 5 hours. Part d: Graph the function on a coordinate plane with an appropriate scale. Part e: Use the graph to estimate the total cost for 3.5 hours of service. Real-World Linear Function: Writing &amp; Interpreting a Graph This lesson builds a complete linear function from a real-world pricing scenario, connecting the equation C(h) = 12h + 40 to its slope, y-intercept, input-output table, and graph. Students practice writing, interpreting, and using a linear function in slope-intercept form to make predictions. In this lesson, you will learn to: • Write a</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/HNoYqTf8ENg</video:player_loc>
      <video:publication_date>2026-07-27T18:48:27.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/plotting-points-and-identifying-quadrants-on-the-coordinate-plane-ezglpcldnli</loc>
    <lastmod>2026-07-27</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/eZgLPclDNLI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Plotting Points and Identifying Quadrants on the Coordinate Plane</video:title>
      <video:description>Question: Plot and label each point on a coordinate plane. Then identify the quadrant or axis where each point is located. Part a: A at 3 comma 5. Part b: B at negative 4 comma 2. Part c: C at negative 1 comma negative 6. Part d: D at 7 comma negative 3. Part e: E at 0 comma 4. Part f: F at negative 5 comma 0. Plotting Points and Identifying Quadrants on the Coordinate Plane This lesson covers how to plot ordered pairs on a coordinate plane and identify whether each point lies in Quadrant I, II, III, or IV, or on one of the axes. Students learn to use the signs of the x- and y-coordinates as a quick guide to location. In this lesson, you will learn to: • Plot ordered pairs accurately on a coordinate plane by moving horizontally then vertically from the origin. • Identify the quadrant of a point by examining the signs of its x- and y-coordinates. • Recognize that any point with a zero</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/eZgLPclDNLI</video:player_loc>
      <video:publication_date>2026-07-27T18:48:31.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/graphing-linear-functions-in-slope-intercept-form-0yl6u-6izvy</loc>
    <lastmod>2026-07-27</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/0yl6U_6IZVY/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Graphing Linear Functions in Slope-Intercept Form</video:title>
      <video:description>Question: Graph each linear function by identifying the slope and y-intercept. Label at least two points on each graph. Part a: y equals three-fourths x plus 1. Part b: y equals negative 2x plus 1. Part c: y equals x minus 4. Part d: y equals negative one-third x minus 2. Graphing Linear Functions in Slope-Intercept Form This lesson covers how to graph linear functions written in slope-intercept form y = mx + b. Students practice identifying the slope and y-intercept from four equations and plotting at least two points to draw each line accurately. In this lesson, you will learn to: • Identify the slope and y-intercept from an equation in slope-intercept form • Plot the y-intercept and use rise-over-run to locate a second point • Graph a linear function by drawing a line through two labeled points • Compare the steepness and direction of lines with positive and negative slopes Course</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/0yl6U_6IZVY</video:player_loc>
      <video:publication_date>2026-07-27T19:08:39.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/determining-whether-a-relation-is-a-function-xcmdkmk19vo</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/xcmdkmK19Vo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Determining Whether a Relation Is a Function</video:title>
      <video:description>Question: A shop tracks the number of items sold each hour during opening hours. The data is shown in the following table. Time in hours after opening, paired with Items Sold: hour 1 sold 24 items, hour 2 sold 35 items, hour 3 sold 35 items, hour 4 sold 48 items, and hour 5 sold 52 items. Determine whether this relation is a function. Explain your reasoning using the definition of a function. Determining Whether a Relation Is a Function This lesson uses a data table of hours and items sold to explore the definition of a function. Students learn to identify inputs and outputs, check for repeated inputs, and apply the definition to decide whether a relation qualifies as a function. Special attention is given to the case where two different inputs share the same output value. In this lesson, you will learn to: • Write ordered pairs from a table and identify the domain and range. • State</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/xcmdkmK19Vo</video:player_loc>
      <video:publication_date>2026-07-29T13:26:52.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/identifying-functions-from-sets-graphs-and-equations-tdgebjrx1oc</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/TdgeBjRX1Oc/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Identifying Functions from Sets, Graphs, and Equations</video:title>
      <video:description>Question: Three relations are shown. Relation A is the set of ordered pairs: negative one comma four, zero comma two, one comma zero, two comma two, and three comma four. Relation B is a graph showing points at negative two comma one, negative one comma three, zero comma two, one comma three, and two comma one. Relation C is the equation x squared plus y equals 9. Part a: For Relation A, determine if it is a function by checking if each input has exactly one output. Part b: For Relation B, apply the vertical line test to determine if it represents a function. Part c: For Relation C, solve for y and explain whether y is a function of x. Identifying Functions from Sets, Graphs, and Equations This lesson examines three representations of relations — a set of ordered pairs, a graph of discrete points, and an equation — and determines whether each one is a function. Students apply the defin</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/TdgeBjRX1Oc</video:player_loc>
      <video:publication_date>2026-07-29T13:26:45.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/evaluating-and-analyzing-a-linear-function-kjt2ol2zbpg</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/kJT2ol2zbPg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Evaluating and Analyzing a Linear Function</video:title>
      <video:description>Question: Given the function f of x equals negative 2x plus 7, where x represents the number of video game rentals per day: Part a — Evaluate f of 3 and explain what it means in context. Part b — If the store can process between 0 and 8 rentals per day, identify the domain and range of this function in context. Part c — Find the value of x when f of x equals 3. Evaluating and Analyzing a Linear Function This lesson explores a linear function in slope-intercept form by evaluating it at a specific input, identifying its domain and range within a real-world constraint, and solving for an unknown input given a specific output. Students practice substitution, algebraic solving, and interpreting function values in context. In this lesson, you will learn to: • Evaluate a linear function at a given input value and interpret the result in context. • Identify the domain and range of a linear fu</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/kJT2ol2zbPg</video:player_loc>
      <video:publication_date>2026-07-29T13:26:44.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/finding-the-linear-function-from-two-points-on-a-graph-rheialn-7zw</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/rHEiaLn-7zw/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Finding the Linear Function from Two Points on a Graph</video:title>
      <video:description>Question: A parking garage charges a base fee plus an hourly rate. After 2 hours, the total charge is 16 dollars. After 5 hours, the total charge is 31 dollars. Write a linear function f of h that represents the total cost in dollars as a function of hours parked h. Then identify the base fee and the hourly rate. Finding the Linear Function from Two Points on a Graph This lesson shows how to build a linear function from two data points using slope-intercept form. Students find the slope and y-intercept from a real-world context, write the function, and verify it against the original data. In this lesson, you will learn to: • Calculate the slope of a linear function using two given points • Determine the y-intercept by substituting the slope and a known point into slope-intercept form • Write and interpret a linear function f(h) = mh + b in a real-world context • Verify a linear functi</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/rHEiaLn-7zw</video:player_loc>
      <video:publication_date>2026-07-29T13:26:36.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/interpreting-linear-functions-in-real-world-context-pikr1k47ny0</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/pikR1K47nY0/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Interpreting Linear Functions in Real-World Context</video:title>
      <video:description>Question: A phone company&apos;s monthly bill is modeled by the function c of m equals 35 plus 0.12 times m, where m is the number of minutes used and c of m is the total cost in dollars. Part a: Identify the y-intercept and explain what it represents. Part b: Find the slope and interpret its meaning. Part c: If a customer uses between 200 and 800 minutes per month, find the domain and range of this function in this context. Interpreting Linear Functions in Real-World Context This lesson models a real-world monthly phone bill using a linear function in slope-intercept form. Students identify and interpret the slope and y-intercept in context, then determine a restricted domain and the corresponding range. In this lesson, you will learn to: • Identify the slope and y-intercept of a linear function written in slope-intercept form and interpret each in context. • Evaluate a linear function at</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/pikR1K47nY0</video:player_loc>
      <video:publication_date>2026-07-29T13:26:31.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/writing-and-comparing-linear-function-rules-bdztjw6nthm</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Bdztjw6NTHM/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Writing and Comparing Linear Function Rules</video:title>
      <video:description>Question: Two fitness centers offer membership plans. Center A charges 40 dollars per month with no enrollment fee. Center B charges 25 dollars per month with a 90 dollar enrollment fee. Part a: Write linear functions f of m and g of m for Centers A and B respectively, where m is the number of months. Part b: Graph both functions on the same coordinate plane. Part c: Find when both centers cost the same amount. Part d: For someone planning to use the gym for 12 months, which center is the better deal and by how much? Writing and Comparing Linear Function Rules This lesson uses a real-world cost-comparison context to build fluency with linear functions in slope-intercept form. Students write, graph, and interpret two linear functions, find their intersection algebraically, and use the break-even point to make a data-driven decision. In this lesson, you will learn to: • Write linear fun</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Bdztjw6NTHM</video:player_loc>
      <video:publication_date>2026-07-29T13:26:30.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/finding-function-values-from-an-equation-bloosnvhpqa</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/BLoosnvHpqA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Finding Function Values from an Equation</video:title>
      <video:description>Question: Let f of x equal negative 2x plus 7. Find f of negative 3, f of 0, and f of 5. Then explain what each result represents as a point on the graph of f. Finding Function Values from an Equation This lesson covers how to evaluate a linear function at specific input values by substituting into the rule f(x) = −2x + 7. Students connect each output to a coordinate point on the graph and interpret key features such as slope, y-intercept, and x-intercept. In this lesson, you will learn to: • Evaluate a linear function f(x) = mx + b at given x-values by substituting and simplifying. • Interpret each function value f(x) as the y-coordinate of a point (x, f(x)) on the graph. • Identify the y-intercept and x-intercept of a linear function from its equation and graph. • Describe how the sign of the slope determines whether a linear function is increasing or decreasing. Course: Algebra1 ·</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/BLoosnvHpqA</video:player_loc>
      <video:publication_date>2026-07-29T20:21:06.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/writing-and-using-a-function-model-zl0ctiwhftw</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/ZL0CtIWhFTw/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Writing and Using a Function Model</video:title>
      <video:description>Question: A cell phone plan charges a 15-dollar monthly fee plus 8 cents per minute of talk time. Write a function C of m that gives the total monthly cost in dollars for m minutes of talk time. Then find C of 200 and C of 500, and interpret what these values mean in context. Writing and Using a Function Model This lesson introduces linear functions by modeling a real-world cost scenario with a fixed fee and a per-unit rate. Students write a linear function in slope-intercept form, evaluate it at specific inputs, graph it, and interpret the results in context. In this lesson, you will learn to: • Write a linear function to model a situation with a fixed fee and a variable rate • Identify the slope and y-intercept of a linear function and explain their real-world meaning • Evaluate a linear function at given input values • Interpret function output values in the context of the original</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/ZL0CtIWhFTw</video:player_loc>
      <video:publication_date>2026-07-29T20:21:01.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-equations-using-function-notation-vhrsajaezna</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/VhrsAJAeznA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Equations Using Function Notation</video:title>
      <video:description>Question: Let g of x equal 3x minus 4. For what value of x does g of x equal 11? Verify your solution and describe the corresponding point on the graph of g. Solving Equations Using Function Notation This lesson demonstrates how to find the input value of a linear function that produces a given output. Students set up and solve a two-step equation, verify the solution by substitution, and interpret the result as a point on the graph of the function. In this lesson, you will learn to: • Set up an equation by substituting a given output value into a linear function • Solve a two-step linear equation using inverse operations • Verify a solution by substituting back into the original function • Interpret the solution as a coordinate point on the graph of the function Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 3- Graphing Linear Functions Level: Grade 9 Pause before each st</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/VhrsAJAeznA</video:player_loc>
      <video:publication_date>2026-07-29T20:20:54.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/converting-standard-form-to-graph-f8mjmafo8au</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/f8MjMAfO8aU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Converting Standard Form to Graph</video:title>
      <video:description>Question: A bakery&apos;s revenue equation is given in standard form: 3x plus 4y equals 24, where x represents the number of loaves sold and y represents the number of pastries sold. Convert this equation to slope-intercept form and graph it on a coordinate plane. Identify the x-intercept and y-intercept, then interpret what each intercept represents in the context of the bakery. Converting Standard Form to Graph This lesson guides students through converting a linear equation from standard form to slope-intercept form, then identifying and graphing the x- and y-intercepts. Students also practice interpreting the slope and intercepts in a real-world context to build conceptual understanding alongside procedural fluency. In this lesson, you will learn to: • Convert a linear equation from standard form (Ax + By = C) to slope-intercept form (y = mx + b). • Identify the x-intercept and y-inter</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/f8MjMAfO8aU</video:player_loc>
      <video:publication_date>2026-07-29T20:20:54.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/standard-form-with-negative-coefficients-rgmwwialiqc</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/rgMwwiAlIqc/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Standard Form with Negative Coefficients</video:title>
      <video:description>Question: Graph the equation 5x minus 2y equals 10 using the standard form method. Find both intercepts algebraically, plot them on a graph, and draw the line. Then determine three points that lie on this line and verify that each one satisfies the original equation. Standard Form with Negative Coefficients This lesson teaches how to graph a linear equation written in standard form by finding the x-intercept and y-intercept algebraically. Students also convert to slope-intercept form to confirm intercepts and verify additional points on the line. In this lesson, you will learn to: • Find the x-intercept and y-intercept of a linear equation in standard form by substituting zero for each variable • Convert a linear equation from standard form to slope-intercept form to identify the slope • Plot intercepts and additional points on a coordinate plane and draw the corresponding line • Veri</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/rgMwwiAlIqc</video:player_loc>
      <video:publication_date>2026-07-29T20:20:52.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/finding-slope-and-y-intercept-from-an-equation-wjaai6bw-o</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/WjAaI6Bw-_o/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Finding Slope and Y-Intercept from an Equation</video:title>
      <video:description>Question: Write the equation y equals negative 3 x plus 7 in slope-intercept form and identify the slope and y-intercept. Then describe what each value tells you about the graph. Finding Slope and Y-Intercept from an Equation This lesson explores the slope-intercept form of a linear equation, $y = mx + b$, using the example $y = -3x + 7$. Students learn to identify the slope and y-intercept directly from the equation, interpret what each value means graphically, and use those values to plot the line accurately. In this lesson, you will learn to: • Recognize and define slope-intercept form $y = mx + b$ • Identify the slope and y-intercept from a linear equation written in slope-intercept form • Interpret the slope as rise over run and explain the direction and steepness of the line • Use the slope and y-intercept to graph a linear equation on the coordinate plane Course: Algebra1 · Bi</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/WjAaI6Bw-_o</video:player_loc>
      <video:publication_date>2026-07-29T20:32:57.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/graphing-using-slope-intercept-form-dce5wncxmmu</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Dce5wNcxMmU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Graphing Using Slope-Intercept Form</video:title>
      <video:description>Question: Graph the equation y equals two-fifths x minus 4 by first plotting the y-intercept, then using the slope to find at least two more points. Verify that all points satisfy the equation. Graphing Using Slope-Intercept Form This lesson covers how to graph a linear equation written in slope-intercept form by identifying the slope and y-intercept, plotting key points using rise-over-run, and verifying that each point satisfies the original equation. Students practice the complete graphing process from equation to plotted line. In this lesson, you will learn to: • Identify the slope and y-intercept from an equation written in slope-intercept form. • Plot the y-intercept and use the slope to locate additional points on the line. • Verify that plotted points satisfy the original linear equation by substitution. • Draw and interpret the graph of a linear equation, including identifyin</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Dce5wNcxMmU</video:player_loc>
      <video:publication_date>2026-07-29T20:33:04.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/converting-standard-form-to-slope-intercept-form-and-graphing-duazafmeem</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/_duAZAfMEEM/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Converting Standard Form to Slope-Intercept Form and Graphing</video:title>
      <video:description>Question: Convert the equation 3x plus 2y equals 12 to slope-intercept form. Then identify the slope and y-intercept, and sketch the graph using these values. Converting Standard Form to Slope-Intercept Form and Graphing This lesson shows how to convert a linear equation from standard form to slope-intercept form by isolating y. Students identify the slope and y-intercept, then use those values to locate key points and sketch the graph. In this lesson, you will learn to: • Convert a linear equation from standard form to slope-intercept form by applying inverse operations. • Identify the slope and y-intercept from the equation y = mx + b. • Find the x-intercept by substituting y = 0 and solving for x. • Graph a linear equation using the slope and intercept points. Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 3- Graphing Linear Functions Level: Grade 9 Pause before each st</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/_duAZAfMEEM</video:player_loc>
      <video:publication_date>2026-07-29T20:33:08.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/direct-variation-recipe-scaling-problem-ejewrist558</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/EJEWRIst558/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Direct Variation: Recipe Scaling Problem</video:title>
      <video:description>Question: A recipe for trail mix requires 3 cups of nuts for every 2 cups of dried fruit. Write a direct variation equation relating the cups of nuts, n, to the cups of dried fruit, d. Then determine how many cups of nuts are needed if you use 8 cups of dried fruit. Direct Variation: Recipe Scaling Problem This lesson introduces direct variation equations of the form n = k·d, where k is the constant of variation. Students learn to identify k from a given ratio, write the equation, graph the relationship, and use it to solve for an unknown quantity. In this lesson, you will learn to: • Identify the constant of variation from a given ratio and write a direct variation equation • Graph a direct variation equation as a line through the origin and interpret its slope • Substitute a known value into a direct variation equation to solve for an unknown Course: Algebra1 · Big Ideas Math · Ron</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/EJEWRIst558</video:player_loc>
      <video:publication_date>2026-07-29T20:51:09.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/direct-variation-hourly-earnings-model-fr-glvy9vvw</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/FR_glVy9Vvw/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Direct Variation: Hourly Earnings Model</video:title>
      <video:description>Question: A worker earns 14 dollars and 50 cents per hour. Write a direct variation equation where earnings, e, vary directly with hours worked, h. If the worker needs to earn 232 dollars, how many hours must they work? Show your work and verify your answer. Direct Variation: Hourly Earnings Model This lesson introduces direct variation equations of the form e = k·h, where k is the constant of variation (unit rate). Students write a direct variation equation from a given rate, solve for an unknown quantity, verify the solution, and interpret the relationship graphically. In this lesson, you will learn to: • Write a direct variation equation by identifying the constant of variation from a unit rate. • Solve a direct variation equation for an unknown variable using division. • Verify a solution by substituting it back into the original equation. • Interpret the graph of a direct variati</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/FR_glVy9Vvw</video:player_loc>
      <video:publication_date>2026-07-29T20:51:12.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/direct-variation-distance-speed-relationship-ztncqna0s5o</loc>
    <lastmod>2026-07-29</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/zTnCQna0s5o/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Direct Variation: Distance-Speed Relationship</video:title>
      <video:description>Question: Two cars travel at constant speeds. Car A travels 156 miles in 3 hours, and Car B travels 117 miles in 2.25 hours. Write a direct variation equation for each car relating distance d to time t. Which car is faster? At their respective constant speeds, how much farther will the faster car travel in 5 hours compared to the slower car? Direct Variation: Distance-Speed Relationship This lesson uses two cars traveling at constant speeds to build direct variation equations of the form d = k·t, where k is the constant of variation (speed in mph). Students compute k for each car, compare speeds, graph the equations, and calculate distances over a given time interval. In this lesson, you will learn to: • Write a direct variation equation $d = k \cdot t$ by computing the constant of variation $k = \dfrac{d}{t}$ from given data. • Compare constants of variation to determine which relati</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/zTnCQna0s5o</video:player_loc>
      <video:publication_date>2026-07-29T20:51:16.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/vertical-shift-of-a-linear-function-nrsm9kxaqe8</loc>
    <lastmod>2026-07-30</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/nRSm9kxaQe8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Vertical Shift of a Linear Function</video:title>
      <video:description>Question: The graph of f of x equals 2x is shifted up 5 units to create a new function g of x. Write the equation for g of x and identify the y-intercept of the new function. Then describe how the slope is affected by this transformation. Vertical Shift of a Linear Function This lesson explores what happens to a linear function when its graph is shifted vertically. Students learn how adding a constant to a function moves the y-intercept while leaving the slope completely unchanged. In this lesson, you will learn to: • Write the equation of a vertically translated linear function by adding a constant to the parent function. • Identify the new y-intercept after a vertical translation. • Explain why a vertical translation changes the y-intercept but does not affect the slope. • Graph both the parent function and the translated function and compare them. Course: Algebra1 · Big Ideas Math</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/nRSm9kxaQe8</video:player_loc>
      <video:publication_date>2026-07-30T10:03:16.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/horizontal-shift-and-reflection-combined-fyzh2puq80q</loc>
    <lastmod>2026-07-30</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/FyZh2pUq80Q/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Horizontal Shift and Reflection Combined</video:title>
      <video:description>Question: Start with the parent function f of x equals x. First shift the graph 3 units to the right, then reflect it across the x-axis to create function h of x. Write the equation for h of x. Verify your equation by testing the point where the original function crosses the y-axis and where the transformed function crosses the y-axis. Horizontal Shift and Reflection Combined This lesson explores how to build a transformed linear function by applying a horizontal shift and an x-axis reflection to a parent function. Students write the equation of the resulting function and verify it by tracking key points through each transformation step. In this lesson, you will learn to: • Apply a horizontal shift to a linear parent function by replacing x with (x − c). • Apply a reflection across the x-axis by multiplying a function by −1. • Write the equation of a transformed linear function in slo</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/FyZh2pUq80Q</video:player_loc>
      <video:publication_date>2026-07-30T10:03:22.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/multiple-transformations-and-function-composition-cmcfmqcd1mc</loc>
    <lastmod>2026-07-30</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/cMCFmQcd1mc/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Multiple Transformations and Function Composition</video:title>
      <video:description>Question: The linear function p of x equals negative 3x plus 2 undergoes the following transformations in order: first, shift left 4 units; second, stretch vertically by a factor of 2; third, shift down 6 units. Write the equation of the final transformed function q of x. Find three points on the original function and verify where they map to on the transformed function. Multiple Transformations and Function Composition This lesson applies three sequential transformations — a horizontal shift left, a vertical stretch, and a vertical shift down — to a linear function. Students derive the equation of the final function step by step and verify the result by tracing three specific points through each transformation. In this lesson, you will learn to: • Apply horizontal and vertical shifts and vertical stretches to a linear function one step at a time • Derive a single combined transformat</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/cMCFmQcd1mc</video:player_loc>
      <video:publication_date>2026-07-30T10:03:27.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/multi-step-inequality-word-problem-61ifzxu2-is</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/61IFZXU2-Is/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Multi-Step Inequality Word Problem</video:title>
      <video:description>Question: A club is selling candles for a fundraiser. Each candle sells for 8 dollars. The club already has 35 dollars donated. They need to raise at least 200 dollars total to fund their event. Write and solve a multi-step inequality to find the minimum number of candles c the club must sell. Interpret the solution in a complete sentence. Multi-Step Inequality Word Problem This lesson walks through writing and solving a multi-step linear inequality from a real-world fundraising context. Students learn to translate &apos;at least&apos; language into an inequality, isolate the variable using inverse operations, and interpret the solution — including the need to round up to a whole number — in context. In this lesson, you will learn to: • Translate a real-world &apos;at least&apos; condition into a multi-step linear inequality • Solve a two-step linear inequality by applying inverse operations in the corre</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/61IFZXU2-Is</video:player_loc>
      <video:publication_date>2026-07-26T22:49:59.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/writing-and-graphing-linear-inequalities-rhhl9omudjo</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/RhhL9OMUDJo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Writing and Graphing Linear Inequalities</video:title>
      <video:description>Question: A roller coaster requires riders to be at least 48 inches tall. Part a: Write an inequality that represents the height h, in inches, a person must be to ride. Part b: Graph the inequality on a number line. Part c: Is a person who is 51 inches tall allowed to ride? Is a person who is 47 inches tall allowed to ride? Explain. Writing and Graphing Linear Inequalities This lesson covers how to translate a real-world &apos;at least&apos; condition into a linear inequality, graph the solution set on a number line using a closed circle and shaded ray, and verify whether specific values satisfy the inequality. Students practice interpreting inequality language and checking solutions by substitution. In this lesson, you will learn to: • Translate phrases such as &apos;at least&apos; into the correct inequality symbol (≥) • Write a linear inequality to represent a real-world constraint • Graph a linear in</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/RhhL9OMUDJo</video:player_loc>
      <video:publication_date>2026-07-26T22:49:51.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-one-step-linear-inequalities-6vy0vuuthiw</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/6VY0VUUThiw/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving One-Step Linear Inequalities</video:title>
      <video:description>Question: Solve each inequality and graph the solution on a number line. Part a: 4x is greater than 28. Part b: n divided by 3 is less than or equal to negative 5. Part c: 12 is less than 6t. Also explain why the inequality sign keeps the same direction when dividing by a positive number. Solving One-Step Linear Inequalities This lesson covers solving one-step linear inequalities by multiplying or dividing both sides by a positive number. Students learn why the inequality sign does not change direction when dividing by a positive number, and how to represent solutions on a number line using open and closed circles. In this lesson, you will learn to: • Solve one-step linear inequalities by dividing or multiplying both sides by a positive number • Explain why the inequality sign stays the same when dividing by a positive number • Represent the solution set of an inequality on a number l</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/6VY0VUUThiw</video:player_loc>
      <video:publication_date>2026-07-26T22:49:47.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-and-graphing-linear-inequalities-in-context-xe0o5bgy9c</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/_XE0O5Bgy9c/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving and Graphing Linear Inequalities in Context</video:title>
      <video:description>Question: A savings account currently holds 45 dollars. The goal is to have more than 120 dollars saved before making a purchase. Write and solve an inequality to find the minimum additional amount d, in dollars, that must be saved. Graph the solution and interpret it in context. Solving and Graphing Linear Inequalities in Context This lesson walks through writing and solving a one-variable linear inequality based on a real-world savings scenario. Students learn to isolate the variable, graph the solution on a number line with an open circle, and interpret what the solution means in context. In this lesson, you will learn to: • Write a linear inequality from a real-world context • Solve a one-step linear inequality by applying inverse operations • Graph the solution set on a number line, correctly using open or closed circles • Interpret and verify the solution in the context of the o</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/_XE0O5Bgy9c</video:player_loc>
      <video:publication_date>2026-07-26T22:49:42.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-and-graphing-linear-inequalities-ut3y37h4gcq</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Ut3y37h4GCQ/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving and Graphing Linear Inequalities</video:title>
      <video:description>Question: Solve each inequality and graph the solution on a number line. Part a: 3x plus 7 is greater than 19. Part b: 5 minus 2n is less than or equal to 13. Part c: 4 times the quantity y minus 3 is greater than or equal to negative 8. For part c, use two methods: first, distribute and then solve; second, divide both sides by 4 first and then solve. Confirm both methods give the same answer. Solving and Graphing Linear Inequalities This lesson covers solving one- and two-step linear inequalities, including cases that require distributing and cases that require dividing by a negative number. Students learn the critical rule about flipping the inequality sign and how to represent solutions on a number line using open and closed circles. In this lesson, you will learn to: • Solve linear inequalities using addition, subtraction, multiplication, and division, applying the sign-flip rule</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Ut3y37h4GCQ</video:player_loc>
      <video:publication_date>2026-07-26T22:49:32.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-linear-inequalities-using-addition-and-subtraction-d-3rezi9jqa</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/d-3RezI9JQA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Linear Inequalities Using Addition and Subtraction</video:title>
      <video:description>Question: Solve each inequality and graph the solution on a number line. Part a: x plus 9 is less than 14. Part b: m minus 6 is greater than or equal to negative 2. Part c: 7 plus y is less than or equal to 3. For part c, a student claims the solution is y is less than or equal to 10. Find and explain the student&apos;s error, then give the correct solution. Solving Linear Inequalities Using Addition and Subtraction This lesson covers solving one-step linear inequalities by applying the Addition and Subtraction Property of Inequalities, graphing solutions on a number line, and identifying common errors. Students practice three inequalities and analyze a worked example containing a mistake to deepen conceptual understanding. In this lesson, you will learn to: • Apply the Addition and Subtraction Property of Inequalities to isolate a variable and solve one-step inequalities • Graph solution</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/d-3RezI9JQA</video:player_loc>
      <video:publication_date>2026-07-26T22:49:37.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
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  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-compound-inequalities-ylnw4talisq</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Ylnw4TAlISQ/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Compound Inequalities</video:title>
      <video:description>Question: Solve each compound inequality and graph the solution on a number line. Part a: negative 4 is less than 2x plus 6, which is less than or equal to 14. Part b: 3 is less than or equal to 5 minus n, which is less than 9. For part a, also describe the solution in words, write it in interval notation, and create a real-world scenario modeled by the inequality. Solving Compound Inequalities This lesson covers how to solve &apos;and&apos; compound inequalities written in three-part form by applying arithmetic operations to all three parts simultaneously. Students learn to write solutions in interval notation, describe them in words, and graph them on a number line, with special attention to flipping inequality symbols when multiplying or dividing by a negative number. In this lesson, you will learn to: • Solve a three-part compound inequality by isolating the variable through operations appl</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Ylnw4TAlISQ</video:player_loc>
      <video:publication_date>2026-07-26T22:49:26.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/reading-and-writing-inequalities-from-number-line-graphs-ttspaoxkfag</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/ttSPaoXKfAg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Reading and Writing Inequalities from Number-Line Graphs</video:title>
      <video:description>Question: Write the inequality shown by each graph and describe the solution in words. Part (a): a number line with an open circle at negative 3 and shading to the left. Part (b): a number line with a closed circle at 5 and shading to the right. Part (c): write a real-world situation that could be modeled by the inequality in part (b). Reading and Writing Inequalities from Number-Line Graphs This lesson teaches students how to translate number-line graphs into algebraic inequalities using open and closed circles and the direction of shading. Students also practice connecting inequalities to real-world situations. In this lesson, you will learn to: • Distinguish between open and closed circles on a number line and connect them to strict vs. non-strict inequalities • Write the correct inequality symbol based on the direction of shading on a number line • Describe the solution set of an</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/ttSPaoXKfAg</video:player_loc>
      <video:publication_date>2026-07-26T22:49:55.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-a-one-step-linear-inequality-add-or-subtract-vxunppe4pve</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/vXUNppe4PVE/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving a One-Step Linear Inequality | Add or Subtract</video:title>
      <video:description>Question: Solve the inequality and graph the solution on a number line. x minus 7 is greater than negative 3. Write the solution in inequality notation and interval notation. Solving a One-Step Linear Inequality | Add or Subtract This lesson walks through solving a one-step linear inequality by applying the Addition Property of Inequality. Students learn to isolate the variable, express the solution in both inequality and interval notation, graph the solution on a number line, and verify correctness using test values. In this lesson, you will learn to: • Solve a one-step linear inequality by adding the same value to both sides • Express the solution in both inequality notation and interval notation • Graph the solution set on a number line using open or closed circles and shading • Verify a solution by substituting test values into the original inequality Course: Algebra1 · Big Ideas</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/vXUNppe4PVE</video:player_loc>
      <video:publication_date>2026-07-26T22:49:21.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
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  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-a-one-step-linear-inequality-multiply-or-divide-sxuuctjqxr4</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/sxuUctJQXR4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving a One-Step Linear Inequality | Multiply or Divide</video:title>
      <video:description>Question: Solve the inequality and graph the solution on a number line. Negative four x is greater than or equal to twenty. Write the solution in inequality notation and interval notation. Explain what happens to the inequality symbol when you divide by a negative number. Solving a One-Step Linear Inequality | Multiply or Divide This lesson covers solving a one-step linear inequality that requires dividing by a negative number. Students learn the critical rule that dividing or multiplying both sides of an inequality by a negative number reverses the inequality symbol, and practice expressing solutions in inequality notation, interval notation, and on a number line. In this lesson, you will learn to: • Solve a linear inequality by isolating the variable using division • Apply the rule that dividing by a negative number reverses the inequality symbol • Express the solution in both inequ</video:description>
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      <video:publication_date>2026-07-26T22:49:17.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-a-multi-step-linear-inequality-jkafvkf66ia</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/JkaFvKF66IA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving a Multi-Step Linear Inequality</video:title>
      <video:description>Question: Solve the inequality, graph the solution, and write the solution in interval notation. 3x plus 8 is less than or equal to 2x minus 4. Check your solution by substituting a value from the solution set back into the original inequality. Solving a Multi-Step Linear Inequality This lesson walks through solving a two-step linear inequality by applying the Addition and Subtraction Properties of Inequality. Students learn to isolate the variable, express the solution in interval notation, graph it on a number line, and verify the answer by substituting a test value. In this lesson, you will learn to: • Solve a linear inequality in one variable using inverse operations • Express the solution set in interval notation using correct bracket and parenthesis notation • Graph the solution set on a number line using open and closed dots • Verify a solution by substituting a test value into</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/JkaFvKF66IA</video:player_loc>
      <video:publication_date>2026-07-26T22:49:13.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-a-compound-inequality-and-70h9meoteyo</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/70h9meOteYo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving a Compound Inequality | AND</video:title>
      <video:description>Question: Solve the compound inequality and graph the solution on a number line: negative 2 is less than 3x minus 5, which is less than or equal to 10. Write the solution in inequality notation and interval notation. State whether the endpoints are included or excluded. Solving a Compound Inequality | AND This lesson covers how to solve a three-part compound (&quot;and&quot;) inequality by applying inverse operations to all three parts simultaneously. Students will express the solution in both inequality notation and interval notation, and verify the answer using test values. In this lesson, you will learn to: • Solve a three-part compound inequality by performing the same operation on all three parts simultaneously • Express the solution in both inequality notation and interval notation • Determine whether each endpoint is included or excluded based on the inequality symbol • Verify the soluti</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/70h9meOteYo</video:player_loc>
      <video:publication_date>2026-07-26T22:49:09.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-a-real-world-linear-inequality-word-problem-ecxarhekzro</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/ECXARhEkZRo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving a Real-World Linear Inequality | Word Problem</video:title>
      <video:description>Question: A fundraiser sells candles for 6 dollars each. The club has already raised 45 dollars from donations. They need to raise MORE THAN 150 dollars in total. Part a: Write an inequality to represent the situation, where c is the number of candles sold. Part b: Solve the inequality. Part c: What is the minimum number of whole candles the club must sell? Explain your reasoning. Part d: Graph the solution on a number line. Solving a Real-World Linear Inequality | Word Problem This lesson teaches students how to write, solve, and graph a one-variable linear inequality derived from a real-world fundraising scenario. Students practice translating verbal conditions into algebraic inequalities, applying inverse operations while tracking the inequality symbol, and interpreting continuous solutions in a whole-number context. In this lesson, you will learn to: • Translate a real-world situa</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/ECXARhEkZRo</video:player_loc>
      <video:publication_date>2026-07-26T22:49:05.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-one-step-equations-multiplication-division-iyqui3cjegy</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/IyquI3cjegY/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving One-Step Equations | Multiplication &amp; Division</video:title>
      <video:description>Question: Solve each equation and check your solutions. Part a: 6 times m equals negative 54. Part b: n divided by 8 equals negative 3. Part c: negative 5 times p equals 35. Solving One-Step Equations | Multiplication &amp; Division This lesson covers solving one-step equations where a variable is multiplied or divided by a constant. Students learn to apply inverse operations — division to undo multiplication, and multiplication to undo division — and then verify each solution by substituting it back into the original equation. In this lesson, you will learn to: • Apply inverse operations to isolate a variable in one-step multiplication and division equations • Solve equations with positive and negative coefficients or divisors • Check solutions by substituting the found value back into the original equation • Recognize that multiplying or dividing by a negative number changes the sign of</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/IyquI3cjegY</video:player_loc>
      <video:publication_date>2026-07-26T11:13:05.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-one-step-equations-addition-subtraction-fus6zggfdjq</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/fUS6ZGgfdjQ/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving One-Step Equations | Addition &amp; Subtraction</video:title>
      <video:description>Question: Solve each equation and check the solution. Part a: x plus 14 equals 31. Part b: y minus 9 equals negative 4. Part c: negative 7 plus z equals 22. Solving One-Step Equations | Addition &amp; Subtraction This lesson covers solving one-step linear equations that involve addition and subtraction. Students learn to isolate the variable by applying inverse operations to both sides of the equation, then verify each solution by substitution. In this lesson, you will learn to: • Apply inverse operations (addition and subtraction) to isolate a variable in a one-step equation • Maintain equality by performing the same operation on both sides of an equation • Use the commutative property of addition to rewrite equations in a clearer form • Verify solutions by substituting back into the original equation Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1 Level: Grade 8 Pause befor</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/fUS6ZGgfdjQ</video:player_loc>
      <video:publication_date>2026-07-26T11:12:45.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-multi-step-equations-two-operations-1len2rb5lia</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/1lEN2RB5LIA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Multi-Step Equations | Two Operations</video:title>
      <video:description>Question: Solve each equation and check your solutions. Part a: 3x plus 7 equals 28. Part b: 4y minus 11 equals 13. Part c: negative 2z plus 5 equals negative 9. Solving Multi-Step Equations | Two Operations This lesson covers solving two-step linear equations of the form ax + b = c by applying inverse operations in reverse order. Students practice subtracting or adding a constant first, then dividing by the coefficient, and verify each solution by substitution. In this lesson, you will learn to: • Apply inverse operations in reverse order to isolate a variable in a two-step equation • Solve two-step linear equations involving positive and negative coefficients • Verify solutions by substituting back into the original equation • Recognize the general solution pattern x = (c − b) / a for equations of the form ax + b = c Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1 Level:</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/1lEN2RB5LIA</video:player_loc>
      <video:publication_date>2026-07-26T11:13:00.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-multi-step-equations-with-the-distributive-property-yezgzjn-du4</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/yeZgZJN_Du4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Multi-Step Equations with the Distributive Property</video:title>
      <video:description>Question: Solve each equation and check your solution. Part a: 2 times the quantity 3x minus 4, equals 16. Part b: negative 3 times the quantity y plus 6, equals 9. Part c: 5 times the quantity 2w minus 1, plus 8, equals 33. Solving Multi-Step Equations with the Distributive Property This lesson teaches students how to solve multi-step linear equations that require applying the distributive property before isolating the variable. Three worked examples progress from basic distribution to combining like terms, and each solution is verified by substitution. In this lesson, you will learn to: • Apply the distributive property to expand expressions within equations • Combine like terms on the same side of an equation • Isolate the variable using inverse operations in the correct order • Verify solutions by substituting back into the original equation Course: Algebra 1 — Solving Multi-Step</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/yeZgZJN_Du4</video:player_loc>
      <video:publication_date>2026-07-26T11:12:21.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/rewriting-literal-equations-solving-for-a-variable-lxpjvsp3iug</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/lXPJvsp3iUg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Rewriting Literal Equations | Solving for a Variable</video:title>
      <video:description>Question: Solve each equation for the indicated variable. Part a: P equals 2l plus 2w, solve for w. Part b: A equals one-half times b times h, solve for b. Part c: V equals l times w times h, solve for l. Part d: C equals 2 pi r, solve for r. Rewriting Literal Equations | Solving for a Variable This lesson covers how to rearrange common geometric formulas — perimeter, area, volume, and circumference — to isolate a specified variable. Students apply inverse operations (addition/subtraction and multiplication/division) to literal equations containing multiple variables, treating all non-target variables as constants. In this lesson, you will learn to: • Identify the target variable in a literal equation and apply inverse operations to isolate it • Rearrange multi-step literal equations involving addition, subtraction, multiplication, and fractions • Solve four standard geometric formula</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/lXPJvsp3iUg</video:player_loc>
      <video:publication_date>2026-07-26T11:12:53.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/absolute-value-equations-basic-cases-pntg0by87ew</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/pntG0BY87ew/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Absolute Value Equations | Basic Cases</video:title>
      <video:description>Question: Solve each equation and check both solutions in the original equation. Part a: the absolute value of x equals 11. Part b: the absolute value of the quantity m minus 4 equals 7. Part c: the absolute value of the quantity 3n plus 6 equals 15. Absolute Value Equations | Basic Cases This lesson covers how to solve absolute value equations by splitting them into two linear cases. Each part is solved step by step and both solutions are verified by substituting back into the original equation. In this lesson, you will learn to: • Apply the split-case property of absolute value to write two linear equations from one absolute value equation • Solve two-step linear equations that arise from absolute value problems • Check both solutions by substituting them back into the original absolute value equation • Recognize that every absolute value equation of the form |expression| = a yields</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/pntG0BY87ew</video:player_loc>
      <video:publication_date>2026-07-26T11:12:49.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/absolute-value-equations-multi-step-no-solution-cj6me1phoao</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Cj6me1phoAo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Absolute Value Equations | Multi-Step &amp; No Solution</video:title>
      <video:description>Question: Solve each equation, or state if there is no solution. Check all solutions. Part a: 2 times the absolute value of x plus 3, minus 5, equals 9. Part b: the absolute value of 4y minus 8, plus 6, equals 2. Part c: 3 times the absolute value of 2z minus 1, plus 4, equals 19. Absolute Value Equations | Multi-Step &amp; No Solution This lesson teaches students how to solve absolute value equations by first isolating the absolute value expression, then splitting into two linear cases. Special attention is given to recognizing when an equation yields no solution because the isolated absolute value equals a negative number. In this lesson, you will learn to: • Isolate an absolute value expression using inverse operations • Apply the definition of absolute value to split one equation into two linear equations • Identify equations with no solution when the isolated absolute value equals a</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Cj6me1phoAo</video:player_loc>
      <video:publication_date>2026-07-26T11:12:39.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/special-solutions-no-solution-and-identity-equations-mwmejgqzve4</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/MwMEJgqZvE4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Special Solutions: No Solution and Identity Equations</video:title>
      <video:description>Question: Solve each equation and classify the result. Part a: 4 times the quantity x plus 2, equals 4x minus 6. Part b: 3 times the quantity 2y minus 1, equals 6y minus 3. Part c: 5z plus 10 equals 5 times the quantity z plus 3. Explain what each result means about the number of solutions. Special Solutions: No Solution and Identity Equations This lesson explores linear equations that produce special outcomes — no solution (contradiction) or infinitely many solutions (identity). Students practice distributing, combining like terms, and interpreting the resulting statement to classify the number of solutions. In this lesson, you will learn to: • Distribute and simplify both sides of a linear equation • Recognize when a variable cancels to produce a false or true statement • Classify a linear equation as having no solution or infinitely many solutions • Explain the meaning of a contrad</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/MwMEJgqZvE4</video:player_loc>
      <video:publication_date>2026-07-26T11:12:26.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/writing-and-solving-one-variable-equations-from-word-problems-cdvtcqyrvg</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/_CDVTCqYRVg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Writing and Solving One-Variable Equations from Word Problems</video:title>
      <video:description>Question: Set up and solve an equation for each scenario. Part a: A phone plan charges a 25-dollar flat fee plus 10 cents per text message. If the bill is 47 dollars, how many texts were sent? Part b: Two hikers start at opposite ends of a 24-mile trail and walk toward each other at 3 miles per hour and 5 miles per hour respectively. How many hours until they meet? Part c: A rectangular garden has a perimeter of 64 feet. The length is 4 feet more than twice the width. Find the dimensions. Writing and Solving One-Variable Equations from Word Problems This lesson demonstrates how to translate three classic real-world scenarios — a phone bill, a meeting-distance problem, and a rectangle perimeter problem — into one-variable equations. Students practice identifying unknowns, building equations from verbal descriptions, and solving using inverse operations. In this lesson, you will learn t</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/_CDVTCqYRVg</video:player_loc>
      <video:publication_date>2026-07-26T11:12:30.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-linear-equations-with-variables-on-both-sides-yyoj-nd0aai</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/yyoJ-nD0aAI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Linear Equations with Variables on Both Sides</video:title>
      <video:description>Question: Solve each equation. Part a: 5x plus 3 equals 2x plus 18. Part b: 7 minus 4y equals 2y minus 11. Part c: 6 times the quantity w minus 2, equals 3w plus 6. Identify the solution, or state if it has no solution or infinitely many solutions. Solving Linear Equations with Variables on Both Sides This lesson covers how to solve linear equations where the variable appears on both sides, including equations requiring the distributive property. Students learn to collect variable terms on one side, isolate the variable, and identify whether an equation has one solution, no solution, or infinitely many solutions. In this lesson, you will learn to: • Collect variable terms on one side of an equation using addition or subtraction properties of equality • Apply the distributive property before solving equations with parentheses • Solve linear equations with variables on both sides and ve</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/yyoJ-nD0aAI</video:player_loc>
      <video:publication_date>2026-07-26T11:12:34.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-one-step-multiplication-division-equations-v2lrhb4v3vi</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/V2lrhb4v3vI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving One-Step Multiplication &amp; Division Equations</video:title>
      <video:description>Question: Solve the equation: 7x equals 91. Show each step and verify your solution. Solving One-Step Multiplication &amp; Division Equations This lesson demonstrates how to solve a one-step linear equation in which the variable is multiplied by a constant. Students learn to apply the Division Property of Equality to isolate the variable and then verify the solution by substitution. In this lesson, you will learn to: • Apply the Division Property of Equality to isolate a variable in a one-step equation • Simplify both sides of an equation to find the value of the unknown • Verify a solution by substituting it back into the original equation Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1- Solving Linear Equations Level: Grade 9 Pause before each step, try the next part yourself, and then continue to check your reasoning. Find more learning support: https://doassignment.ca T</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/V2lrhb4v3vI</video:player_loc>
      <video:publication_date>2026-07-26T15:51:10.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-a-literal-equation-for-a-specific-variable-ce7dpoxldua</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/ce7dPOxlduA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving a Literal Equation for a Specific Variable</video:title>
      <video:description>Question: The formula for the perimeter of a rectangle is P equals 2 times l plus 2 times w. Part a: Solve the formula for l, the length. Part b: Use the result to find the length of a rectangle with a perimeter of 54 centimeters and a width of 9 centimeters. Solving a Literal Equation for a Specific Variable This lesson teaches students how to rearrange a multi-variable formula by isolating a specific variable using inverse operations. Using the rectangle perimeter formula P = 2l + 2w, students solve for length and then apply the rearranged formula to a numerical problem. In this lesson, you will learn to: • Rearrange a literal equation to isolate a specified variable using inverse operations • Apply a rearranged formula by substituting given values to find an unknown quantity • Verify a solution by substituting back into the original formula Course: Algebra1 · Big Ideas Math · Ron</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/ce7dPOxlduA</video:player_loc>
      <video:publication_date>2026-07-26T15:45:09.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-equations-with-fractions-ebyyvdxeusa</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/ebyYvdXeUSA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Equations with Fractions</video:title>
      <video:description>Question: Solve the equation: three-fourths x plus 2 equals one-half x plus 7. Eliminate fractions using the least common denominator, then solve. Verify your answer. Solving Equations with Fractions This lesson teaches how to solve a linear equation containing fractions by using the least common denominator to clear all fractions, then applying inverse operations to isolate the variable. Students also learn how to verify their solution by substituting back into the original equation. In this lesson, you will learn to: • Identify the least common denominator of fractions appearing in a linear equation • Eliminate fractions by multiplying every term on both sides by the LCD • Isolate the variable using inverse operations to find the solution • Verify a solution by substituting it back into the original equation Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1- Solving Linear</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/ebyYvdXeUSA</video:player_loc>
      <video:publication_date>2026-07-26T15:50:32.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-multi-step-equations-with-like-terms-etjauknfa4e</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/EtjaUKNfa4E/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Multi-Step Equations with Like Terms</video:title>
      <video:description>Question: Solve the equation: 5x plus 3x minus 10 equals 38. Combine like terms first, then isolate the variable. Show all steps and check your solution. Solving Multi-Step Equations with Like Terms This lesson walks through solving a linear equation that requires combining like terms before isolating the variable. Students learn to apply the Addition and Division Properties of Equality in sequence, then verify their solution by substitution. In this lesson, you will learn to: • Identify and combine like terms to simplify a linear equation • Apply the Addition and Division Properties of Equality to isolate a variable • Verify a solution by substituting it back into the original equation • Communicate each algebraic step clearly and in logical order Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1- Solving Linear Equations Level: Grade 8 Pause before each step, try the next</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/EtjaUKNfa4E</video:player_loc>
      <video:publication_date>2026-07-26T15:50:36.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-equations-using-the-distributive-property-qjap8wtwags</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/QjAP8wTwAgs/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Equations Using the Distributive Property</video:title>
      <video:description>Question: Solve the equation: 4 times the quantity 2x minus 3, equals 28. Apply the distributive property first, then solve the resulting two-step equation. Check your answer. Solving Equations Using the Distributive Property This lesson walks through solving a linear equation that requires the distributive property before isolating the variable. Students learn to expand parentheses, then apply inverse operations in sequence, and verify the solution by substitution. In this lesson, you will learn to: • Apply the distributive property to expand expressions with parentheses • Solve a two-step linear equation using inverse operations • Check a solution by substituting it back into the original equation Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1- Solving Linear Equations Level: Grade 9 Pause before each step, try the next part yourself, and then continue to check your re</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/QjAP8wTwAgs</video:player_loc>
      <video:publication_date>2026-07-26T15:50:41.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-absolute-value-equations-szoagnmwdng</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/szOaGnMwDng/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Absolute Value Equations</video:title>
      <video:description>Question: Solve the equation: the absolute value of 2x minus 6 equals 10. Write two separate cases, solve each, and check both solutions in the original equation. Solving Absolute Value Equations This lesson covers how to solve an absolute value equation of the form |ax + b| = c by splitting it into two separate linear cases. Students solve each case independently and verify both solutions by substituting back into the original equation. In this lesson, you will learn to: • Explain why an absolute value equation produces two separate cases • Write and solve each linear case resulting from an absolute value equation • Check both solutions by substituting them into the original equation • State the complete solution set for an absolute value equation Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1- Solving Linear Equations Level: Grade 9 Pause before each step, try the next</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/szOaGnMwDng</video:player_loc>
      <video:publication_date>2026-07-26T15:50:45.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/writing-solving-a-linear-equation-from-a-word-problem-sjivsrwvmjk</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/SjivsRWvMjk/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Writing &amp; Solving a Linear Equation from a Word Problem</video:title>
      <video:description>Question: A movie streaming service charges a flat monthly fee of 8 dollars plus 3 dollars per premium movie rented. If a customer&apos;s bill for one month is 26 dollars, write and solve an equation to find how many premium movies the customer rented. Writing &amp; Solving a Linear Equation from a Word Problem This lesson teaches students how to translate a real-world billing scenario into a two-step linear equation and solve for the unknown. Students practice isolating the variable using inverse operations — subtraction and division — and verify their solution by substitution. In this lesson, you will learn to: • Translate a word problem into a two-step linear equation • Solve a two-step equation using inverse operations • Verify the solution by substituting it back into the original equation Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1- Solving Linear Equations Level: Grade 9</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/SjivsRWvMjk</video:player_loc>
      <video:publication_date>2026-07-26T15:50:49.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-equations-with-variables-on-both-sides-1cod0a5olag</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/1COD0A5oLAg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Equations with Variables on Both Sides</video:title>
      <video:description>Question: Solve the equation: 6x plus 5 equals 2x plus 21. Collect variable terms on one side and constants on the other. Show each step and verify the solution. Solving Equations with Variables on Both Sides This lesson teaches students how to solve a two-step linear equation by collecting variable terms on one side and constant terms on the other through inverse operations. Each algebraic step is shown explicitly, and the solution is verified by substitution into the original equation. In this lesson, you will learn to: • Collect variable terms on one side of an equation by applying inverse operations • Isolate the variable by undoing addition and division in the correct order • Verify a solution by substituting it back into the original equation Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1- Solving Linear Equations Level: Grade 9 Pause before each step, try the next</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/1COD0A5oLAg</video:player_loc>
      <video:publication_date>2026-07-26T15:50:54.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-two-step-equations-crgkxmojdfo</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/CRgkxMOJdFo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving Two-Step Equations</video:title>
      <video:description>Question: Solve the equation: 3x minus 8 equals 16. Identify the two operations applied to x and undo them in reverse order. Verify your answer. Solving Two-Step Equations This lesson teaches students how to solve a two-step linear equation by identifying the operations applied to the variable and undoing them in reverse order. Students practice isolating the variable through addition and division, then verify their solution by substitution. In this lesson, you will learn to: • Identify the two operations applied to a variable in a linear equation • Apply inverse operations in reverse order to isolate the variable • Solve a two-step linear equation algebraically • Verify a solution by substituting it back into the original equation Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1- Solving Linear Equations Level: Grade 9 Pause before each step, try the next part yourself, a</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/CRgkxMOJdFo</video:player_loc>
      <video:publication_date>2026-07-26T15:51:00.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/solving-one-step-addition-subtraction-equations-nxq7kmbg4jq</loc>
    <lastmod>2026-07-26</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/NXq7kmBG4jQ/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Solving One-Step Addition &amp; Subtraction Equations</video:title>
      <video:description>Question: Solve the equation: x plus 14 equals 29. Show each step and check your answer by substituting back into the original equation. Solving One-Step Addition &amp; Subtraction Equations This lesson demonstrates how to solve a one-step linear equation of the form x + a = b by applying the Subtraction Property of Equality. Students learn to isolate the variable by performing the same operation on both sides, then verify their solution by substituting back into the original equation. In this lesson, you will learn to: • Apply the Subtraction Property of Equality to isolate a variable in a one-step equation • Simplify both sides of an equation after performing inverse operations • Verify a solution by substituting it back into the original equation • Communicate each algebraic step clearly and in the correct order Course: Algebra1 · Big Ideas Math · Ron Larson · Chapter 1- Solving Linea</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/NXq7kmBG4jQ</video:player_loc>
      <video:publication_date>2026-07-26T15:51:07.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-16-4-engineering-mechanics-dynamics-hibbeler-14th-edition-lqw3c1-0e88</loc>
    <lastmod>2026-07-25</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/lQW3c1_0E88/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 16-4: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A disk is originally rotating at omega-zero equals 12 radians per second. It is subjected to a constant angular acceleration of alpha equals 20 radians per second squared. Determine the magnitudes of the velocity and the normal and tangential components of acceleration of point B when the disk has undergone 2 revolutions. Point B is located at a radius of 0.4 meters from the center. Problem 16-4: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) Planar Kinematics of a Rotating Disk: Velocity and Acceleration of a Point This lesson applies rotational kinematics to find the angular velocity of a disk after a given angular displacement, then uses the rigid-body relationships to determine the speed and the normal and tangential acceleration components of a point on the disk. The method combines the kinematic equation $\omega^2 = \omega_0^2 + 2\alpha\theta$ with the point-a</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/lQW3c1_0E88</video:player_loc>
      <video:publication_date>2026-07-25T21:52:57.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-16-5-engineering-mechanics-dynamics-hibbeler-14th-edition-ifmqtyv2bha</loc>
    <lastmod>2026-07-25</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/IFMQtYv2BHA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 16-5: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A disk is driven by a motor such that the angular position of the disk is defined by theta equals 20t plus 4t-squared radians, where t is in seconds. Determine the number of revolutions, the angular velocity, and the angular acceleration of the disk when t equals 90 seconds. Problem 16-5: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the calculus-based kinematic relationships for a rigid body in pure rotation. Given an angular position function as a polynomial in time, students find the angular velocity and acceleration by differentiation, then convert angular position to revolutions. In this lesson, you will learn to: • Differentiate a polynomial angular position function to obtain angular velocity and angular acceleration • Evaluate angular kinematic quantities at a specified instant in time • Convert angular displacement from radians to revol</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/IFMQtYv2BHA</video:player_loc>
      <video:publication_date>2026-07-25T21:53:02.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-16-6-engineering-mechanics-dynamics-hibbeler-14th-edition-ikmwbz9akyw</loc>
    <lastmod>2026-07-25</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/IKmwBZ9Akyw/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 16-6: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A wheel has an initial clockwise angular velocity of 10 radians per second and a constant angular acceleration of 3 radians per second squared. Determine the number of revolutions it must undergo to acquire a clockwise angular velocity of 15 radians per second. What time is required? Problem 16-6: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the constant-angular-acceleration kinematic equations to a rotating wheel. Students find the angular displacement (in revolutions) and elapsed time needed for the wheel to reach a target angular velocity, then verify the result using a third kinematic equation. In this lesson, you will learn to: • Apply constant-angular-acceleration equations to relate angular velocity, angular displacement, and time • Convert angular displacement from radians to revolutions • Select the most efficient kinematic equation fo</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/IKmwBZ9Akyw</video:player_loc>
      <video:publication_date>2026-07-25T21:53:07.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-16-7-engineering-mechanics-dynamics-hibbeler-14th-edition-p5fmpmrcv0k</loc>
    <lastmod>2026-07-25</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/p5fmpmrcv0k/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 16-7: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: Gear A rotates with a constant angular acceleration of alpha A equals 90 radians per second squared, starting from rest. Determine the time required for gear D to attain an angular velocity of 600 rpm. Also find the number of revolutions of gear D to attain this angular velocity. Gears A, B, C, and D have radii of 15 millimeters, 50 millimeters, 25 millimeters, and 75 millimeters, respectively. Gears B and C are on the same shaft. Gear A meshes with gear B, and gear C meshes with gear D. Problem 16-7: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson analyzes a two-stage compound gear train (A–B and C–D) under constant angular acceleration. Students learn to propagate angular acceleration through meshing gears and a shared shaft, then apply constant-acceleration kinematics to find the time and number of revolutions required for the output gear to reach a tar</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/p5fmpmrcv0k</video:player_loc>
      <video:publication_date>2026-07-25T21:53:11.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-16-8-engineering-mechanics-dynamics-hibbeler-14th-edition-k3pkoidfbw</loc>
    <lastmod>2026-07-25</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/_k3PkoIDFbw/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 16-8: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: Gear A rotates with an angular velocity of omega A equals theta A plus 1, in radians per second, where theta A is the angular displacement of gear A in radians. Determine the angular acceleration of gear D when theta A equals 3 radians, starting from rest. Gears A, B, C, and D have radii r A equals 15 millimeters, r B equals 50 millimeters, r C equals 25 millimeters, and r D equals 75 millimeters, respectively. Gears B and C are on the same shaft. Problem 16-8: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson analyzes a two-stage gear train (A–B/C–D) where the driving gear has a position-dependent angular velocity. Students apply the kinematic chain-rule relation α = ω dω/dθ to find angular acceleration, then propagate it through both gear stages using contact-velocity conditions. In this lesson, you will learn to: • Apply the kinematic relation α = ω dω/d</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/_k3PkoIDFbw</video:player_loc>
      <video:publication_date>2026-07-25T21:53:15.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-16-9-engineering-mechanics-dynamics-hibbeler-14th-edition-l1xdfokzlcy</loc>
    <lastmod>2026-07-25</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/L1xdfoKzlcY/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 16-9: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: At the instant omega-A equals 5 radians per second, pulley A is given an angular acceleration alpha-A equals 0.8 times theta, in radians per second squared, where theta is in radians. Pulley A has a radius of 50 millimeters and is connected by a belt to the outer rim of pulley C, which has an outer radius of 40 millimeters. Pulley C has an inner hub rigidly fixed to it, with a hub radius of 60 millimeters. Point B lies on the rim of this hub. Determine the magnitude of the acceleration of point B when pulley A has completed 3 full revolutions from the initial instant. Problem 16-9: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson analyzes a two-pulley belt drive system where the driving pulley has a position-dependent angular acceleration. Using kinematic integration and the no-slip belt constraint, we determine the angular motion of the driven pulley and t</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/L1xdfoKzlcY</video:player_loc>
      <video:publication_date>2026-07-25T21:53:19.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-16-10-engineering-mechanics-dynamics-hibbeler-14th-edition-z9hgc5tnkuu</loc>
    <lastmod>2026-07-25</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Z9hgc5TnkuU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 16-10: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: At the instant omega-A equals 5 radians per second, pulley A is given a constant angular acceleration alpha-A equal to 6 radians per second squared. Determine the magnitude of acceleration of point B on pulley C when A rotates 2 full revolutions. Pulley A has a radius of 50 millimeters. Pulley C has an outer radius of 40 millimeters and an inner hub radius of 60 millimeters, where point B lies on the outer rim of the hub. The belt connects the outer rim of A to the outer rim of C. Problem 16-10: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson analyzes a two-pulley belt system under constant angular acceleration. Starting from given initial conditions, students find the angular velocity after a specified rotation, transfer motion through the belt constraint, and compute the total acceleration of a point on the compound pulley&apos;s hub. In this lesson, you wil</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Z9hgc5TnkuU</video:player_loc>
      <video:publication_date>2026-07-25T21:53:24.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-15-1-engineering-mechanics-dynamics-hibbeler-14th-edition-7bctjdgh2a4</loc>
    <lastmod>2026-07-23</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/7bCTjDgH2A4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 15-1: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 150-gram ball is kicked such that it leaves the ground at an angle of 60 degrees and strikes the ground at the same elevation a distance of 12 meters away. Determine the impulse of the foot on the ball. Neglect the impulse caused by the ball&apos;s weight while it is being kicked. Problem 15-1: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the principle of impulse and momentum to a projectile problem. Students determine the launch speed of a ball from its known range and launch angle, then use the impulse-momentum theorem to find the magnitude and direction of the impulsive force. The weight impulse is neglected during the very short impact interval. In this lesson, you will learn to: • Apply the principle of linear impulse and momentum to relate an impulsive force to the change in momentum of an object • Use projectile kinematics to determine the</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/7bCTjDgH2A4</video:player_loc>
      <video:publication_date>2026-07-23T13:14:16.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-15-2-engineering-mechanics-dynamics-hibbeler-14th-edition-snpinqpnfae</loc>
    <lastmod>2026-07-23</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/sNpiNqpnFAE/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 15-2: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 20-pound block slides down a 30-degree inclined plane with an initial velocity of 2 feet per second. Determine the velocity of the block after 3 seconds if the coefficient of kinetic friction between the block and the plane is mu-k equals 0.25. Problem 15-2: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the principle of linear impulse and momentum to find the velocity of a block sliding down a rough inclined plane after a given time interval. The method systematically resolves weight components, calculates the normal and friction forces, and uses the impulse-momentum equation along the slope. A kinematic verification confirms the result. In this lesson, you will learn to: • Resolve weight into components parallel and perpendicular to an inclined surface and determine the normal force. • Calculate the kinetic friction force and the net impulse</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/sNpiNqpnFAE</video:player_loc>
      <video:publication_date>2026-07-23T13:15:46.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-15-3-engineering-mechanics-dynamics-hibbeler-14th-edition-bywscyy2ppi</loc>
    <lastmod>2026-07-23</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/bywSCYY2ppI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 15-3: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A uniform beam has a weight of 5000 pounds. Determine the average tension in each of the two cables A B and A C if the beam is given an upward speed of 8 feet per second in 1.5 seconds starting from rest. Neglect the mass of the cables. Hook point A is 4 feet above the beam ends B and C, and B and C are each 3 feet horizontally from the center of the beam. Problem 15-3: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the linear impulse–momentum principle to a beam lifted from rest by two symmetric cables attached at a common hook point. Students find the average cable tension by relating the net vertical impulse to the change in linear momentum, then verify the result using Newton&apos;s second law. In this lesson, you will learn to: • Construct a free-body diagram identifying cable tension components and weight for a symmetric two-cable lifting system</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/bywSCYY2ppI</video:player_loc>
      <video:publication_date>2026-07-23T13:15:59.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-15-4-engineering-mechanics-dynamics-hibbeler-14th-edition-d3j783ch7xy</loc>
    <lastmod>2026-07-23</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/D3j783CH7XY/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 15-4: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: Each of two symmetric cables can sustain a maximum tension of 5000 pounds. A uniform beam weighs 5000 pounds. Hook A is located 4 feet above the beam attachment points B and C, which are each 3 feet from the center of the beam. Determine the shortest time possible to lift the beam from rest to a speed of 10 feet per second. Problem 15-4: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson determines the minimum time to accelerate a uniform beam from rest to a target speed using two symmetric cables at their maximum allowable tension. The solution combines cable geometry, vertical force equilibrium, and the linear impulse-momentum theorem in US customary units. In this lesson, you will learn to: • Resolve cable tensions into vertical and horizontal components using geometry and direction cosines • Apply vertical force equilibrium to find the maximum net upward</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/D3j783CH7XY</video:player_loc>
      <video:publication_date>2026-07-23T13:16:04.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-15-5-engineering-mechanics-dynamics-hibbeler-14th-edition-2owrrhjayqo</loc>
    <lastmod>2026-07-23</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/2oWrRHJayQo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 15-5: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A hockey puck is traveling to the left with a velocity of v1 equals 10 meters per second when it is struck by a hockey stick and given a velocity of v2 equals 20 meters per second directed at 40 degrees above the horizontal, to the upper right. Determine the magnitude of the net impulse exerted by the hockey stick on the puck. The puck has a mass of 0.2 kilograms. Problem 15-5: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the impulse-momentum theorem to a two-dimensional collision between a hockey stick and puck. Students decompose initial and final velocity vectors into components, compute the change in momentum, and find the magnitude and direction of the net impulse vector. In this lesson, you will learn to: • Apply the vector form of the impulse-momentum principle: $\vec{J} = m\vec{v}_2 - m\vec{v}_1$ • Decompose velocity vectors into x- and</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/2oWrRHJayQo</video:player_loc>
      <video:publication_date>2026-07-23T13:43:28.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-15-6-engineering-mechanics-dynamics-hibbeler-14th-edition-wm6grclgwes</loc>
    <lastmod>2026-07-23</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/wm6gRClGWes/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 15-6: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A train consists of a 50-megagram engine and three cars, each having a mass of 30 megagrams. It takes 80 seconds for the train to increase its speed uniformly from rest to 40 kilometers per hour. The engine wheels provide a resultant frictional tractive force F that drives the train forward, while the car wheels roll freely. Determine, first, the tractive force F acting on the engine wheels, and second, the tension T developed at the coupling between the engine E and the first car A. Problem 15-6: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the linear impulse–momentum theorem to a multi-body train system to find both the tractive force at the drive wheels and the internal coupling tension between vehicles. Students learn to isolate subsystems strategically—first the entire train, then a subset of cars—to solve for unknown forces without computi</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/wm6gRClGWes</video:player_loc>
      <video:publication_date>2026-07-23T13:43:32.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-15-7-engineering-mechanics-dynamics-hibbeler-14th-edition-xzg64syk2ba</loc>
    <lastmod>2026-07-23</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/xzg64SYK2BA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 15-7: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: Crate A weighs 100 pounds and crate B weighs 50 pounds. A horizontal force P equal to 50 pounds is applied to crate A, pushing it into crate B so both crates slide together on the ground. The coefficient of kinetic friction between the crates and the ground is mu sub k equal to 0.25. If the crates start from rest, determine: first, their common speed when t equals 5 seconds, and second, the force exerted by crate A on crate B during the motion. Problem 15-7: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the linear impulse-momentum principle to a system of two crates pushed across a rough floor by a horizontal force. Students find the common speed of the system after a given time interval, then isolate one body to determine the internal contact force between the crates. In this lesson, you will learn to: • Apply the principle of linear impulse an</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/xzg64SYK2BA</video:player_loc>
      <video:publication_date>2026-07-23T13:24:36.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-15-8-engineering-mechanics-dynamics-hibbeler-14th-edition-v2gheh-akc8</loc>
    <lastmod>2026-07-23</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/V2ghEH_AkC8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 15-8: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: An automobile has a weight of 2700 pounds and is traveling forward at 4 feet per second when it crashes into a wall. The impact occurs in 0.06 seconds. Part A: Determine the average impulsive force acting on the car when the brakes are NOT applied. Part B: If the coefficient of kinetic friction between the wheels and the pavement is mu sub k equals 0.3, calculate the impulsive force on the wall if the brakes WERE applied during the crash, with all four wheels slipping. Assume the car comes to rest in both cases. Problem 15-8: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the Principle of Linear Impulse and Momentum to determine the average impulsive force a wall exerts on a car during a short-duration collision. Two cases are compared: brakes not applied (wall force only) and brakes applied with all wheels slipping (wall force plus kinetic fricti</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/V2ghEH_AkC8</video:player_loc>
      <video:publication_date>2026-07-23T14:29:53.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-15-9-engineering-mechanics-dynamics-hibbeler-14th-edition-b07docoyrvu</loc>
    <lastmod>2026-07-23</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/B07dOcoYRVU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 15-9: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 200-kilogram crate rests on the ground. The coefficients of static and kinetic friction are mu-s equals 0.5 and mu-k equals 0.4. A winch delivers a horizontal towing force T to its cable, which varies as T equals 400 times t to the one-half, in newtons, while T is less than or equal to 800 newtons, after which T remains constant at 800 newtons. The cable is doubled, meaning two cable segments pull the crate, so the net horizontal force on the crate is 2T. Initially the tension is zero. Determine the speed of the crate at t equals 4 seconds. Problem 15-9: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the impulse-momentum principle to a crate pulled by a doubled cable whose tension varies as a square-root function of time. Students determine when static friction is overcome, set up and evaluate the impulse integral during the sliding phase, and c</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/B07dOcoYRVU</video:player_loc>
      <video:publication_date>2026-07-23T14:40:46.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-15-10-engineering-mechanics-dynamics-hibbeler-14th-edition-lomxf6gus-4</loc>
    <lastmod>2026-07-23</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/LoMXF6guS-4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 15-10: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 50-kilogram crate is pulled from rest by a constant force P applied at 30 degrees above the horizontal. The crate reaches a speed of 10 meters per second in 5 seconds. The coefficient of kinetic friction between the crate and the ground is mu-k equals 0.2. Determine the magnitude of P. Problem 15-10: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the linear impulse-momentum principle to find an unknown applied force on a crate pulled from rest across a rough horizontal surface. The method combines vertical equilibrium (to find the normal force) with the horizontal impulse-momentum equation, accounting for a force applied at an angle and kinetic friction. In this lesson, you will learn to: • Apply the linear impulse-momentum principle to a particle under constant forces over a known time interval • Determine the normal force when an angled appli</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/LoMXF6guS-4</video:player_loc>
      <video:publication_date>2026-07-23T14:29:59.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-14-1-engineering-mechanics-dynamics-hibbeler-14th-edition-kqz1rugm2pi</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/kqZ1ruGM2pI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 14-1: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 20-kilogram crate is subjected to a force having a constant direction and a magnitude F equal to 100 newtons, applied at 30 degrees above the horizontal. When s equals 15 meters, the crate is moving to the right with a speed of 8 meters per second. Determine its speed when s equals 25 meters. The coefficient of kinetic friction between the crate and the ground is mu-sub-k equal to 0.25. Problem 14-1: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the Principle of Work and Energy to find the final speed of a crate acted on by an angled constant force and kinetic friction over a given displacement. Students learn to compute the normal force accounting for the vertical component of the applied force, calculate work done by each force, and solve for the unknown speed using the work–energy equation. In this lesson, you will learn to: • Apply the Wor</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/kqZ1ruGM2pI</video:player_loc>
      <video:publication_date>2026-07-22T18:26:22.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-14-2-engineering-mechanics-dynamics-hibbeler-14th-edition-cp-hyewpxpm</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/cP-HyewPxPM/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 14-2: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A barrel barrier is placed in front of a bridge pier. The force-deflection relation of the barrier is F equals 90 times 10 to the 3rd power, times x to the one-half power, in pounds, where x is in feet. A car weighing 4000 pounds is traveling at 75 feet per second just before it hits the barrier. Determine the car&apos;s maximum penetration into the barrier. Problem 14-2: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the Principle of Work and Energy to find the maximum deflection of a nonlinear barrier struck by a moving vehicle. Students integrate a variable force–deflection relationship and set the resulting work equal to the initial kinetic energy to solve for the unknown penetration depth. In this lesson, you will learn to: • Apply the Work–Energy Theorem to a particle brought to rest by a variable force • Integrate a power-law force function to</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/cP-HyewPxPM</video:player_loc>
      <video:publication_date>2026-07-22T18:26:27.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-14-3-engineering-mechanics-dynamics-hibbeler-14th-edition-qfrx4g5kfji</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/QFRx4G5kFJI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 14-3: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A crate with a mass of 100 kilograms is subjected to two forces: an 800 newton force directed 30 degrees below the horizontal, pushing to the right and downward, and a 1000 newton rope force directed at an angle whose slope components are 3 vertical and 4 horizontal — that is, at the arctangent of 3 over 4 above the horizontal — pulling to the right and upward. The crate starts from rest on a horizontal surface with a coefficient of kinetic friction mu sub k equal to 0.2. Determine the distance s the crate must slide to attain a speed of 6 meters per second. Problem 14-3: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the work-energy theorem to find the distance a crate must travel to reach a target speed when subjected to two angled forces and kinetic friction. Students resolve forces into components, compute the normal and friction forces, then</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/QFRx4G5kFJI</video:player_loc>
      <video:publication_date>2026-07-22T18:26:33.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-14-4-engineering-mechanics-dynamics-hibbeler-14th-edition-ncezpwzjeym</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/NCeZpWzjEYM/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 14-4: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 100-kilogram crate is subjected to two forces: a 500-newton force directed at 45 degrees above the horizontal, pulling upward and to the left, and a 400-newton force directed at 30 degrees below the horizontal, pushing inward from the right. The crate is originally at rest on a horizontal surface. Determine the distance s it must slide in order to attain a speed of v equals 8 meters per second. The coefficient of kinetic friction between the crate and the surface is mu sub k equals 0.2. Use the work-energy principle. Problem 14-4: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the work-energy principle to a crate acted on by two applied forces at different angles and kinetic friction on a horizontal surface. Students resolve each force into components, compute the normal force, determine friction, and use the work-energy theorem to find the dist</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/NCeZpWzjEYM</video:player_loc>
      <video:publication_date>2026-07-22T18:26:39.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-14-5-engineering-mechanics-dynamics-hibbeler-14th-edition-4cm6yxzapmg</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/4Cm6yxZaPMg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 14-5: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A roller coaster car is essentially at rest at the crest of hill A. Determine the required height h so that the car reaches a speed of 100 kilometres per hour at the bottom of hill B. Also determine the minimum radius of curvature rho for the track at B so that passengers do not experience a normal force greater than 4 times m g. Neglect the size of the car and passengers. Problem 14-5: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the work–energy principle to find the height needed for a roller coaster car to reach a target speed, then uses Newton&apos;s second law in the normal direction at the bottom of a curved track to determine the minimum radius of curvature that keeps the normal force within a specified limit. Both conservation of energy and centripetal acceleration concepts are combined in a single two-part problem. In this lesson, you will</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/4Cm6yxZaPMg</video:player_loc>
      <video:publication_date>2026-07-22T18:26:44.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-14-6-engineering-mechanics-dynamics-hibbeler-14th-edition-yjyobq-neyy</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/yJYobQ-neyY/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 14-6: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: When the driver applies the brakes of a light truck traveling 40 kilometers per hour, it skids 3 meters before stopping. How far will the truck skid if it is traveling 80 kilometers per hour when the brakes are applied? Problem 14-6: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the work-energy theorem to determine how a vehicle&apos;s skid distance changes with initial speed. Students learn that because kinetic energy depends on the square of velocity, doubling the speed quadruples the stopping distance under constant friction conditions. In this lesson, you will learn to: • Apply the work-energy theorem to a particle decelerating under a constant friction force • Derive the relationship between initial speed and skid distance by cancelling mass • Use proportional reasoning to predict skid distance when initial speed changes • Verify results numeric</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/yJYobQ-neyY</video:player_loc>
      <video:publication_date>2026-07-22T18:59:47.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-14-7-engineering-mechanics-dynamics-hibbeler-14th-edition-gdl-2gtxezi</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/gdl-2GtXEzI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 14-7: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 10-kilogram block rests on a smooth surface and is subjected to a horizontal force of 6 newtons. Observer A is in a fixed frame x. The block has an initial speed of 5 meters per second and travels 10 meters, both directed to the right and measured from the fixed frame. Observer B moves at a constant velocity of 2 meters per second to the right relative to A, attached to the x-prime axis. Part one: using the principle of work and energy, determine the final speed of the block as measured by Observer A. Part two: repeat using Observer B&apos;s frame, and show that both observers obtain the same final speed for the block relative to their own frame, shifted by 2 meters per second, thereby demonstrating that the work–energy principle is valid in any inertial reference frame. Problem 14-7: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the principle of wo</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/gdl-2GtXEzI</video:player_loc>
      <video:publication_date>2026-07-22T18:59:53.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-14-8-engineering-mechanics-dynamics-hibbeler-14th-edition-m34hptpnxuw</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/M34hpTpNxuw/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 14-8: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A force F equal to 250 newtons is applied at point B of a pulley-rope system. Determine the speed of the 10-kilogram block at A when it has moved 1.5 meters upward, starting from rest. The pulley arrangement gives the rope length constraint: S-W plus 2 times s-F equals l, a constant, where S-W is the position of the block and s-F is the position of point B where the force F is applied. Problem 14-8: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the work-energy theorem to a pulley-rope system in which a 250 N applied force lifts a 10 kg block. Students first use the rope length constraint to relate displacements at each end, then compute net work done by all active forces, and finally solve for the block&apos;s speed after it rises 1.5 m from rest. In this lesson, you will learn to: • Derive the kinematic relationship between displacements in a pulley</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/M34hpTpNxuw</video:player_loc>
      <video:publication_date>2026-07-22T19:00:39.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-14-9-engineering-mechanics-dynamics-hibbeler-14th-edition-s0pqkk7ozbk</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/s0pqKk7ozBk/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 14-9: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: An air spring is used to protect a support and prevent damage to a tensioning weight in the event of a belt failure. The force developed by the air spring as a function of its deflection is linear: the force increases from 0 Newtons at s equals 0 meters, to 1500 Newtons at s equals 0.2 meters. A block of mass 20 kilograms is suspended a height d equals 0.4 meters above the top of the undeformed spring. Determine the maximum deformation s-max of the spring when the belt fails. Neglect the mass of the pulley and belt. Problem 14-9: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the work-energy theorem to find the maximum compression of a linear air spring that catches a falling mass after a sudden load release. Students integrate a linearly varying spring force and solve the resulting quadratic equation to determine the maximum deflection, verifying</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/s0pqKk7ozBk</video:player_loc>
      <video:publication_date>2026-07-22T19:00:44.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-14-10-engineering-mechanics-dynamics-hibbeler-14th-edition-1jjt5fdjpu0</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/1jjt5fdJPU0/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 14-10: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A horizontal force acting on a 20-kilogram block has a magnitude that varies with position s according to F equals 50 times the square root of s, in newtons. When s equals zero, the block is moving to the right at v-sub-1 equals 6 meters per second. The coefficient of kinetic friction between the block and the surface is 0.3. Determine the distance s-sub-2 the block must slide before its velocity reaches v-sub-2 equals 15 meters per second. Problem 14-10: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the Work–Energy Theorem to find the displacement of a block acted on by a position-dependent force and kinetic friction. The method integrates the variable force to compute work, combines it with the constant friction work, and solves a nonlinear equation numerically for the unknown displacement. In this lesson, you will learn to: • Apply the Work–E</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/1jjt5fdJPU0</video:player_loc>
      <video:publication_date>2026-07-22T19:00:50.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-13-1-engineering-mechanics-dynamics-hibbeler-14th-edition-wf-pyfbt3rm</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/wf-pyFBT3rM/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 13-1: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 6-pound particle is subjected to the action of its weight and three applied forces. Force F-1 equals 2i plus 6j minus 2t k, in pounds. Force F-2 equals t-squared i minus 4t j minus 1k, in pounds. Force F-3 equals negative 2t i, in pounds, where t is in seconds. Determine the distance the particle is from the origin 2 seconds after being released from rest. Problem 13-1: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies Newton&apos;s Second Law in three dimensions to a particle subjected to multiple time-varying forces and gravity. By summing force components, dividing by mass, and integrating twice, students find the particle&apos;s position vector as a function of time and compute its distance from the origin at a specified instant. In this lesson, you will learn to: • Sum multiple time-varying force vectors component-by-component to find the net force. •</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/wf-pyFBT3rM</video:player_loc>
      <video:publication_date>2026-07-22T10:03:53.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-13-2-engineering-mechanics-dynamics-hibbeler-14th-edition-j8eik31gfmk</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/j8EIK31Gfmk/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 13-2: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: Two boxcars A and B have weights of 20,000 pounds and 30,000 pounds, respectively. They are freely coasting down an incline when brakes are applied to all wheels of car A. Determine the force in the coupling C between the two cars. The coefficient of kinetic friction between the wheels of A and the tracks is mu sub k equals 0.5. The wheels of car B are free to roll. Neglect wheel mass. The incline is a 1-in-10 grade, meaning tangent of theta equals 1 over 10, so sine of theta equals 1 over the square root of 101, and cosine of theta equals 10 over the square root of 101. Solve using single resultant normal forces on A and B. Problem 13-2: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies Newton&apos;s Second Law to a two-car system rolling down an inclined track when brakes are applied to one car&apos;s wheels. Students find the common deceleration of the syst</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/j8EIK31Gfmk</video:player_loc>
      <video:publication_date>2026-07-22T10:03:58.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-13-3-engineering-mechanics-dynamics-hibbeler-14th-edition-d-q3jyzb5pu</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/D-q3JyzB5pU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 13-3: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 50-kilogram crate starts from rest on a horizontal surface. A horizontal force P equal to 200 newtons is applied to the crate. The coefficient of kinetic friction between the crate and the ground is mu sub k equal to 0.3. Determine: (a) the distance the crate travels, and (b) its velocity when t equals 3 seconds. Problem 13-3: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies Newton&apos;s second law to find the acceleration of a crate sliding on a surface with kinetic friction. Using constant-acceleration kinematics, students then determine the crate&apos;s displacement and velocity at a given time. In this lesson, you will learn to: • Apply Newton&apos;s second law in the vertical direction to find the normal force on a horizontal surface • Calculate the kinetic friction force and net horizontal force acting on a particle • Use Newton&apos;s second law in the horiz</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/D-q3JyzB5pU</video:player_loc>
      <video:publication_date>2026-07-22T10:04:02.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-13-4-engineering-mechanics-dynamics-hibbeler-14th-edition-3pe16gizmuw</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/3PE16GIZmuw/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 13-4: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 50-kilogram crate starts from rest and achieves a velocity of 4 meters per second when it travels a distance of 5 meters to the right. Determine the magnitude of force P acting on the crate. The force P is applied at an angle of 30 degrees above the horizontal. The coefficient of kinetic friction between the crate and the ground is mu-k equals 0.3. Problem 13-4: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies Newton&apos;s second law to a crate accelerating on a rough horizontal surface under an angled applied force. Students use kinematics to find acceleration, then set up and solve equations of motion in both the vertical and horizontal directions to determine the unknown force magnitude. In this lesson, you will learn to: • Use a kinematic equation to determine constant acceleration from initial velocity, final velocity, and distance. • Construct</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/3PE16GIZmuw</video:player_loc>
      <video:publication_date>2026-07-22T10:04:06.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-13-5-engineering-mechanics-dynamics-hibbeler-14th-edition-fvll9cibuko</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/FVlL9cIbuKo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 13-5: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: Blocks A and B of mass 10 kilograms and 6 kilograms, respectively, are placed on an inclined plane where theta equals 30 degrees, and released from rest. They are connected by a rigid link. The coefficients of kinetic friction between the blocks and the inclined plane are mu sub A equals 0.1 and mu sub B equals 0.3. Neglect the mass of the link. Determine the force developed in the link. Problem 13-5: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies Newton&apos;s Second Law to a system of two blocks connected by a rigid link on a 30° inclined plane with different coefficients of kinetic friction. Students learn to determine the common acceleration of the system and the internal force (tension or compression) developed in the connecting link by analyzing the system as a whole and then isolating individual bodies. In this lesson, you will learn to: • Dete</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/FVlL9cIbuKo</video:player_loc>
      <video:publication_date>2026-07-22T10:04:09.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-13-6-engineering-mechanics-dynamics-hibbeler-14th-edition-sn3dzjfasne</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Sn3DZjfAsNE/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 13-6: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 10-pound block has a speed of 4 feet per second when a force F equal to 8 t-squared pounds is applied horizontally. Determine the velocity of the block when t equals 2 seconds. The coefficient of kinetic friction at the surface is mu-sub-k equal to 0.2. Problem 13-6: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the principle of linear impulse and momentum to a block subjected to a time-varying horizontal force and kinetic friction. Students integrate Newton&apos;s second law directly to find the block&apos;s velocity at a specified time, combining force analysis with calculus-based momentum methods. In this lesson, you will learn to: • Identify and calculate all horizontal and vertical forces acting on a sliding block, including a time-varying applied force and kinetic friction. • Apply the impulse-momentum principle by integrating Newton&apos;s second law</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Sn3DZjfAsNE</video:player_loc>
      <video:publication_date>2026-07-22T10:39:14.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-13-7-engineering-mechanics-dynamics-hibbeler-14th-edition-cjcxmia5f7g</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/CJcXmIa5f7g/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 13-7: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 10-pound block has a speed of 4 feet per second when a horizontal force F equals 8t-squared pounds is applied. Determine the velocity of the block when it has moved s equals 30 feet. The coefficient of kinetic friction is mu-sub-k equals 0.2. Impulse-Momentum with a Time-Dependent Force and Kinetic Friction This lesson solves for the velocity of a block subjected to a time-varying horizontal force and kinetic friction after traveling a specified displacement. Because the applied force is a function of time, the impulse-momentum method is combined with kinematic integration to relate time and position, and the result is verified with the work-energy theorem. In this lesson, you will learn to: • Apply Newton&apos;s second law to write the equation of motion for a block with a time-dependent applied force and kinetic friction. • Integrate the equation of motion to obtain velocity</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/CJcXmIa5f7g</video:player_loc>
      <video:publication_date>2026-07-22T10:39:23.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-13-8-engineering-mechanics-dynamics-hibbeler-14th-edition-hglrkblmta</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/-hGlRkblMTA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 13-8: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A 3500-pound car has its speed plotted over a 30-second time period. From the velocity–time graph: for the interval from zero to ten seconds, the speed increases linearly from zero to 60 feet per second; and for the interval from ten seconds to thirty seconds, the speed increases linearly from 60 to 80 feet per second. Determine and plot the variation of the traction force F needed to cause the motion, assuming a horizontal road with no friction or aerodynamic drag. Problem 13-8: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies Newton&apos;s Second Law to find the traction force required to accelerate a vehicle along a horizontal road when its velocity–time profile is piecewise linear. Students extract constant accelerations from each linear segment, compute the corresponding forces, and describe the resulting piecewise-constant force–time graph. In thi</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/-hGlRkblMTA</video:player_loc>
      <video:publication_date>2026-07-22T17:47:20.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-13-9-engineering-mechanics-dynamics-hibbeler-14th-edition-xnjwqwrgqlg</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/xNjwqwRGQLg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 13-9: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A conveyor belt is moving at 4 meters per second. The coefficient of static friction between the conveyor belt and a 10-kilogram package is mu sub s equals 0.2. Determine the shortest time the belt can stop so that the package does not slide on the belt. Problem 13-9: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies Newton&apos;s Second Law and kinematics to determine the shortest time a conveyor belt can decelerate to rest without a package sliding on it. Students use the static friction limit to find the maximum allowable deceleration, then apply a constant-acceleration equation to solve for the minimum stopping time. In this lesson, you will learn to: • Calculate the maximum static friction force on an object resting on a moving surface • Apply Newton&apos;s Second Law to relate the friction force to the deceleration of a package on a conveyor belt • Use</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/xNjwqwRGQLg</video:player_loc>
      <video:publication_date>2026-07-22T10:39:28.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-13-10-engineering-mechanics-dynamics-hibbeler-14th-edition-g9aidou7tm8</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/G9AIdOu7tM8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 13-10: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A conveyor belt transports packages. Each 10-kilogram package has a coefficient of kinetic friction mu-k equal to 0.15. The conveyor belt is moving at 5 meters per second and then suddenly stops. Determine the distance the package will slide on the belt before coming to rest. Problem 13-10: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies Newton&apos;s Second Law and kinematics to find how far a package slides after a moving conveyor belt suddenly stops. The method is verified independently using the Work-Energy Theorem, reinforcing the connection between force-acceleration and energy approaches in particle kinetics. In this lesson, you will learn to: • Identify the direction of kinetic friction on a package sliding relative to a stationary belt • Apply Newton&apos;s Second Law in both vertical and horizontal directions to find normal force and deceleration</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/G9AIdOu7tM8</video:player_loc>
      <video:publication_date>2026-07-22T10:39:19.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-1-dynamics-hibbeler-qmm6ywwhq0u</loc>
    <lastmod>2023-05-11</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/qMm6YWwHQ0U/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12 1 Dynamics Hibbeler</video:title>
      <video:description>Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m/s^2, where t is in seconds. What is the particle’s velocity when t = 6 s, and what is its position when t = 11 s? Please consider subscribing to my channel for more problem solutions. Your likes, shares, and comments are greatly appreciated and help to promote my channel. Thank you! Kinematics of Particles How to solve Kinematics Problems | Doassignment Engineering Dynamics by Hibbeler 14th Edition</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/qMm6YWwHQ0U</video:player_loc>
      <video:publication_date>2023-05-11T04:54:53.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-9-engineering-mechanics-dynamics-hibbeler-14th-edition-4-wadilloog</loc>
    <lastmod>2026-07-18</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/4-WADILLoog/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-9: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Kinematics of a Particle: Velocity, Position, and Total Distance via Integration This lesson demonstrates how to find a particle&apos;s velocity and position functions by integrating a time-dependent acceleration along a straight line. It also covers how to determine total distance traveled by checking for direction reversals using the velocity function. Question: The acceleration of a particle as it moves along a straight line is given by a equals 2t minus 1, in meters per second squared, where t is in seconds. If s equals 1 meter and v equals 2 meters per second when t equals 0, determine the particle&apos;s velocity and position when t equals 6 seconds. Also determine the total distance the particle travels during this time period. In this lesson, you will learn to: • Integrate a given acceleration function to obtain velocity as a function of time, applying initial conditions. • Integrate th</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/4-WADILLoog</video:player_loc>
      <video:publication_date>2026-07-18T15:01:26.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-10-engineering-mechanics-dynamics-hibbeler-14th-edition-09ebwehxya0</loc>
    <lastmod>2026-07-18</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/09EBWehxya0/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-10: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Finding Particle Velocity by Numerical Integration of a Position-Dependent Acceleration This lesson demonstrates how to find a particle&apos;s velocity at a given position when acceleration is a nonlinear function of position. The kinematic relation v dv = a ds is used to set up a definite integral, which is then evaluated numerically using Simpson&apos;s 1/3 Rule with four subintervals. Question: A particle moves along a straight line with an acceleration of a equals 5 divided by the quantity 3 times s to the one-third power plus s to the five-halves power, in meters per second squared, where s is in meters. Determine the particle&apos;s velocity when s equals 2 meters, given that it starts from rest when s equals 1 meter. Use a numerical method to evaluate the integral. In this lesson, you will learn to: • Apply the kinematic relationship a = v dv/ds to separate variables and set up a definite int</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/09EBWehxya0</video:player_loc>
      <video:publication_date>2026-07-18T19:13:06.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-11-engineering-mechanics-dynamics-hibbeler-14th-edition-svbamfaahsu</loc>
    <lastmod>2026-07-18</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/SvBamfaAhsU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-11: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Average Velocity and Average Speed in Rectilinear Kinematics This lesson covers rectilinear (straight-line) kinematics for a particle that changes direction during its motion. Students learn to distinguish displacement from total path length and apply those concepts to compute average velocity (a vector) and average speed (a scalar) over a given time interval. Question: A particle travels along a straight-line path such that in 4 seconds it moves from an initial position s-A equal to negative 8 meters to a position s-B equal to positive 3 meters. Then in another 5 seconds it moves from s-B to s-C equal to negative 6 meters. Determine the particle&apos;s average velocity and average speed during the 9-second time interval. In this lesson, you will learn to: • Calculate displacement as the net change in position between two points in time • Compute average velocity using total displacement d</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/SvBamfaAhsU</video:player_loc>
      <video:publication_date>2026-07-18T18:13:07.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-12-engineering-mechanics-dynamics-hibbeler-14th-edition-topjzocpanm</loc>
    <lastmod>2026-07-18</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/TopJZocPAnM/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-12: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Rectilinear Kinematics with Constant Acceleration This lesson applies the constant-acceleration kinematic equations to find the time and distance required for a car to change speed along a straight road. Students learn to select the appropriate equation for each unknown and verify results using an independent check equation. Question: A car traveling with an initial speed of 70 kilometers per hour accelerates at a constant rate of 6000 kilometers per hour squared along a straight road. First, how long will it take the car to reach a speed of 120 kilometers per hour? Second, through what distance does the car travel during this time? In this lesson, you will learn to: • Identify the appropriate constant-acceleration kinematic equation based on the given and unknown quantities • Solve for elapsed time using the velocity–time equation $v = v_0 + at$ • Solve for displacement using the vel</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/TopJZocPAnM</video:player_loc>
      <video:publication_date>2026-07-18T18:26:25.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
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  <url>
    <loc>https://doassignment.ca/videos/problem-12-13-engineering-mechanics-dynamics-hibbeler-14th-edition-dm-tstsmy-4</loc>
    <lastmod>2026-07-18</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Dm_TSTSmY_4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-13: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Rectilinear Kinematics: Stopping Distance with Reaction Time This lesson applies rectilinear kinematics to calculate total stopping distance when a driver must first react before braking. Students compare two scenarios with different reaction times but identical initial speed and deceleration, reinforcing how the two-phase stopping model (reaction + braking) is used in real engineering problems. Question: A normal driver has a reaction time of 0.75 seconds before braking, while an impaired driver has a reaction time of 3 seconds. Both are traveling at 30 miles per hour, which equals 44 feet per second, on a straight road. Both cars can decelerate at 2 feet per second squared. Determine the shortest total stopping distance d for each driver from the moment they see a hazard. In this lesson, you will learn to: • Decompose total stopping distance into a constant-velocity reaction phase a</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Dm_TSTSmY_4</video:player_loc>
      <video:publication_date>2026-07-18T18:25:29.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
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  <url>
    <loc>https://doassignment.ca/videos/problem-12-14-engineering-mechanics-dynamics-hibbeler-14th-edition-lnwmuqk1zg4</loc>
    <lastmod>2026-07-18</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/LnwMuQk1ZG4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-14: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Rectilinear Kinematics: Distance, Velocity, and Acceleration This lesson analyzes the motion of a particle moving along a straight line using a cubic position function. Students apply differentiation to find velocity and acceleration, identify direction reversals, and distinguish between total distance traveled, average speed, and average velocity. Question: The position of a particle along a straight-line path is defined by s of t equals t cubed minus 6 t squared minus 15 t plus 7 feet, where t is in seconds. Determine the total distance traveled when t equals 10 seconds. Find the particle&apos;s average velocity, average speed, and the instantaneous velocity and acceleration at t equals 10 seconds. In this lesson, you will learn to: • Differentiate a position function to obtain velocity and acceleration functions. • Identify turning points by setting the velocity function equal to zero.</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/LnwMuQk1ZG4</video:player_loc>
      <video:publication_date>2026-07-18T18:29:42.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-15-engineering-mechanics-dynamics-hibbeler-14th-edition-nmczuhycfsg</loc>
    <lastmod>2026-07-18</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/NMczuHycFsg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-15: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Rectilinear Kinematics with Velocity-Dependent Deceleration This lesson derives velocity and position as explicit functions of time for a particle subjected to a deceleration proportional to the cube of its velocity, $a = -kv^3$. The method uses separation of variables and direct integration of the kinematic differential equations, with verification of initial conditions and the original acceleration law. Question: A particle is moving with a velocity of v-naught when s equals zero and t equals zero. If it is subjected to a deceleration of a equals negative k times v cubed, where k is a constant, determine its velocity and position as functions of time. In this lesson, you will learn to: • Set up and separate the kinematic differential equation $a = dv/dt$ when acceleration depends on velocity. • Apply definite integration with initial conditions to obtain $v(t)$ in closed form. • Use</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/NMczuHycFsg</video:player_loc>
      <video:publication_date>2026-07-18T19:13:01.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-16-engineering-mechanics-dynamics-hibbeler-14th-edition-0k1umuinxgs</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/0K1UMUINXGs/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-16: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Rectilinear Kinematics with Velocity-Dependent Acceleration This lesson covers how to analyze straight-line particle motion when acceleration is expressed as a function of velocity. Students apply separation of variables and definite integration to determine both the total distance traveled and the elapsed time before the particle stops. Question: A particle moves along a straight line with an initial velocity of 6 meters per second. It is subjected to a deceleration equal to negative 1.5 times the square root of v, in meters per second squared, where v is in meters per second. Determine: part a, how far the particle travels before it stops; and part b, how much time this takes. In this lesson, you will learn to: • Set up and apply the kinematic relation a = v dv/ds to find displacement when acceleration depends on velocity • Set up and apply the kinematic relation a = dv/dt to find e</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/0K1UMUINXGs</video:player_loc>
      <video:publication_date>2026-07-19T21:04:22.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-17-engineering-mechanics-dynamics-hibbeler-14th-edition-szp-reb-yxi</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/szp-REB_yxI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-17: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Minimum Following Distance to Avoid Collision (Kinematics) This lesson applies one-dimensional particle kinematics to determine the minimum safe following distance between two vehicles undergoing sequential braking. It combines constant-velocity and constant-deceleration motion phases, and uses the condition of zero relative velocity to identify the critical instant of minimum gap. Question: Car B is traveling a distance d ahead of car A. Both cars are traveling at 60 feet per second when the driver of car B suddenly applies the brakes, causing car B to decelerate at 12 feet per second squared. It takes the driver of car A 0.75 seconds to react, which is a normal driver reaction time. When car A&apos;s driver applies the brakes, car A decelerates at 15 feet per second squared. Determine the minimum distance d between the cars so as to avoid a collision. In this lesson, you will learn to: •</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/szp-REB_yxI</video:player_loc>
      <video:publication_date>2026-07-19T21:29:27.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-18-engineering-mechanics-dynamics-hibbeler-14th-edition-cdb-wfngmdu</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/CDb-WfNgmdU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-18: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Rocket Altitude via Kinematics and Integration This lesson finds the time for a rocket to reach a given altitude when acceleration is a function of position. The method combines the kinematic identity a = v dv/ds with separation of variables, completing the square, and a standard logarithmic integral formula. Question: The acceleration of a rocket traveling upward is given by a equals 6 plus 0.02 s, in meters per second squared, where s is in meters. Determine the time needed for the rocket to reach an altitude of s equals 100 meters. Initially, v equals 0 and s equals 0 when t equals 0. In this lesson, you will learn to: • Apply the kinematic identity $a = v\,\dfrac{dv}{ds}$ to convert a position-dependent acceleration into a separable ODE • Integrate to find velocity as a function of position and set up a time integral • Simplify a radical integrand by completing the square • Evalua</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/CDb-WfNgmdU</video:player_loc>
      <video:publication_date>2026-07-19T21:29:31.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-19-engineering-mechanics-dynamics-hibbeler-14th-edition-djmfaxwwf3a</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/djMFaXWWf3A/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-19: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Rectilinear Motion with Constant Acceleration: Three-Phase Train Problem This lesson analyzes a train journey broken into three phases — uniform acceleration from rest, constant-velocity cruising, and uniform deceleration to a stop. Students apply standard kinematic equations to find the distance covered in each phase and sum them to find the total distance between two stations. Question: A train starts from rest at station A and accelerates at 0.5 meters per second squared for 60 seconds. Afterwards it travels with a constant velocity for 15 minutes. It then decelerates at 1 meter per second squared until it is brought to rest at station B. Determine the distance between the stations. In this lesson, you will learn to: • Identify and separate a multi-phase rectilinear motion problem into distinct constant-acceleration intervals • Apply kinematic equations ($v = v_0 + at$ and $d = v_0</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/djMFaXWWf3A</video:player_loc>
      <video:publication_date>2026-07-19T21:29:36.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-20-engineering-mechanics-dynamics-hibbeler-14th-edition-eajzq-ma4p4</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/eAjZQ-Ma4p4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-20: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Kinematics with Calculus: Position, Distance, and Acceleration This lesson applies integral and differential calculus to analyze the motion of a particle along a straight line. Students integrate a velocity function to obtain position, identify direction changes to compute total distance traveled, and differentiate velocity to find acceleration at a specific instant. Question: The velocity of a particle traveling along a straight line is v equals 3t squared minus 6t, in feet per second, where t is in seconds. If s equals 4 feet when t equals 0, determine the position of the particle when t equals 4 seconds. What is the total distance traveled during the time interval t equals 0 to t equals 4 seconds? Also, what is the acceleration when t equals 2 seconds? In this lesson, you will learn to: • Integrate a given velocity function and apply an initial condition to determine the position f</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/eAjZQ-Ma4p4</video:player_loc>
      <video:publication_date>2026-07-19T22:08:49.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-21-engineering-mechanics-dynamics-hibbeler-14th-edition-3c5dmlp2zc</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/-3C5DMlP2zc/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-21: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Kinematics with Exponential Velocity: Distance and Acceleration This lesson applies integral and differential calculus to a velocity function of the form v(t) = 60(1 − e^(−t)). Students find the total distance traveled over a time interval by integrating velocity, then find instantaneous acceleration by differentiating velocity. Question: A freight train travels at a velocity v equals 60 times the quantity 1 minus e to the negative t, in feet per second, where t is the elapsed time in seconds. Determine the distance traveled in three seconds, and the acceleration at that time. In this lesson, you will learn to: • Compute the distance traveled over a given time interval by integrating a velocity function • Derive the acceleration function by differentiating a given velocity function • Evaluate definite integrals and derivatives involving natural exponential functions • Interpret the ph</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/-3C5DMlP2zc</video:player_loc>
      <video:publication_date>2026-07-19T22:08:53.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-22-engineering-mechanics-dynamics-hibbeler-14th-edition-muqkjbssncy</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/MuqkjBsSncY/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-22: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Rectilinear Motion: Sandbag Released from an Ascending Balloon This lesson applies constant-acceleration kinematics to an object released from a vertically ascending platform. Students determine the impact speed of the released object and the altitude of the ascending platform at the moment of impact by carefully setting up a signed coordinate system and using the standard kinematic equations. Question: A sandbag is dropped from a balloon that is ascending vertically at a constant speed of 6 meters per second. The bag is released with the same upward velocity of 6 meters per second at time equals zero, and it hits the ground when time equals 8 seconds. Determine: first, the speed of the bag as it hits the ground; and second, the altitude of the balloon at time equals 8 seconds. In this lesson, you will learn to: • Set up a signed coordinate system and correctly assign initial conditio</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/MuqkjBsSncY</video:player_loc>
      <video:publication_date>2026-07-19T22:08:59.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-23-engineering-mechanics-dynamics-hibbeler-14th-edition-utg4wa2gl9q</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/utG4Wa2gL9Q/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-23: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Kinematics with Velocity-Dependent Acceleration This lesson solves a particle motion problem where acceleration is given as a function of velocity, $a = -2v$. By treating the kinematic relationships as separable first-order ODEs, students derive position, velocity, and acceleration as explicit functions of time using integration and initial conditions. Question: A particle moves along a straight line with acceleration defined as a equals negative 2v meters per second squared, where v is in meters per second. Given that v equals 20 meters per second when s equals 0 meters and t equals 0 seconds, determine the position, velocity, and acceleration as functions of time. In this lesson, you will learn to: • Set up and solve a first-order separable ODE for velocity when acceleration is a function of velocity • Apply initial conditions to determine constants of integration for both velocity</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/utG4Wa2gL9Q</video:player_loc>
      <video:publication_date>2026-07-19T22:09:03.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-24-engineering-mechanics-dynamics-hibbeler-14th-edition-zaeveq9ydlq</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/zaEVeq9YDLQ/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-24: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Particle Velocity from a Position-Dependent Acceleration This lesson demonstrates how to find a particle&apos;s velocity as a function of position when acceleration is given as a function of position. The method uses the chain-rule kinematic identity a = v dv/ds to convert the problem into a separable ordinary differential equation, which is then integrated with an initial condition. Question: The acceleration of a particle traveling along a straight line is a equals one-fourth times the square root of s, in meters per second squared, where s is in meters. If v equals zero and s equals one meter when t equals zero, determine the particle&apos;s velocity at s equals two meters. In this lesson, you will learn to: • Apply the kinematic identity a = v dv/ds to relate acceleration, velocity, and position without using time. • Separate variables and integrate both sides of a differential equation usi</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/zaEVeq9YDLQ</video:player_loc>
      <video:publication_date>2026-07-19T22:09:07.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-25-engineering-mechanics-dynamics-hibbeler-14th-edition-u7pszrv5p3e</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/u7PSzRv5P3E/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-25: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Terminal Velocity via Separable ODEs This lesson models free-fall motion with air resistance using a separable first-order ODE of the form dv/dt = k[1 − αv²]. Students separate variables, apply partial fractions, integrate, and apply an initial condition to find an explicit velocity function involving hyperbolic tangent, then evaluate it at a specific time and determine the terminal velocity analytically. Question: A falling body has an acceleration defined by a equals 9.81 times the quantity one minus v-squared times ten-to-the-negative-four, in meters per second squared, where v is in meters per second and the positive direction is downward. The body is released from rest at a very high altitude. Determine: part a, the velocity when t equals 5 seconds, and part b, the body&apos;s terminal, or maximum attainable, velocity as t approaches infinity. In this lesson, you will learn to: • Set</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/u7PSzRv5P3E</video:player_loc>
      <video:publication_date>2026-07-19T22:09:11.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-26-engineering-mechanics-dynamics-hibbeler-14th-edition-o0isjav-u-4</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/O0iSJaV_u-4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-26: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Integration of Motion Equations: Position, Distance, and Velocity This lesson demonstrates how to find a particle&apos;s velocity and position functions by integrating a time-dependent acceleration using initial conditions. It then applies those functions to determine total distance traveled by identifying direction reversals through the zeros of velocity. Question: The acceleration of a particle along a straight line is defined by a equals 2t minus 9 meters per second squared, where t is in seconds. At t equals 0, s equals 1 meter and v equals 10 meters per second. When t equals 9 seconds, determine: (a) the particle&apos;s position, (b) the total distance traveled, and (c) the velocity. In this lesson, you will learn to: • Integrate a given acceleration function to obtain velocity, applying an initial condition to determine the constant of integration. • Integrate the velocity function to obt</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/O0iSJaV_u-4</video:player_loc>
      <video:publication_date>2026-07-19T23:39:35.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-27-engineering-mechanics-dynamics-hibbeler-14th-edition-bq5-msne95s</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/bq5-mSNe95s/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-27: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Solving a Separable ODE for Terminal Velocity Kinematics This lesson applies separation of variables and partial fractions to solve a first-order ODE modeling air-resistance drag, where acceleration decreases from g to zero as a particle approaches terminal velocity. Students find the explicit time required to reach a given fraction of terminal velocity starting from rest. Question: A particle falls through air with acceleration given by a equals g over v-f squared, times the quantity v-f squared minus v squared, where g is gravitational acceleration and v-f is the terminal velocity. The particle starts from rest. Determine the time t needed for the velocity to reach v equals v-f divided by 2. In this lesson, you will learn to: • Set up and classify a separable first-order ODE from a physical acceleration model • Apply partial fraction decomposition to integrate a rational function of</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/bq5-mSNe95s</video:player_loc>
      <video:publication_date>2026-07-19T23:39:40.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-28-engineering-mechanics-dynamics-hibbeler-14th-edition-plxyawo3hxc</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/PLXyaWo3Hxc/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-28: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Two Projectiles Passing Each Other: Constant Acceleration This lesson applies the constant-acceleration position equation to two balls launched simultaneously in opposite directions along a vertical line. By setting the two position functions equal, students discover that the shared gravitational term cancels, reducing the problem to a simple linear equation for the meeting time and height. Question: Ball A is thrown vertically upward from the top of a 30-meter-high building with an initial velocity of 5 meters per second. At the same instant, ball B is thrown vertically upward from the ground with an initial velocity of 20 meters per second. Determine the height above the ground and the time at which the two balls pass each other. In this lesson, you will learn to: • Derive position equations for objects under constant gravitational acceleration using integration of kinematics equati</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/PLXyaWo3Hxc</video:player_loc>
      <video:publication_date>2026-07-19T23:39:46.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-29-engineering-mechanics-dynamics-hibbeler-14th-edition-krqwxoyfhi4</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/kRqwXoYfhI4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-29: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Rectilinear Motion with Variable Acceleration This lesson covers how to analyze the motion of two particles moving along a straight line under variable acceleration by integrating acceleration to find velocity and position. Students learn to determine the separation between particles at a given instant and to calculate total distance traveled by identifying direction reversals. Question: Two particles A and B start from rest at the origin s equals 0 and move along a straight line such that the acceleration of A equals 6t minus 3 feet per second squared, and the acceleration of B equals 12t squared minus 8 feet per second squared, where t is in seconds. Determine the distance between them when t equals 4 seconds, and the total distance each has traveled in t equals 4 seconds. In this lesson, you will learn to: • Integrate variable acceleration functions to obtain velocity and position</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/kRqwXoYfhI4</video:player_loc>
      <video:publication_date>2026-07-19T23:39:51.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-30-engineering-mechanics-dynamics-hibbeler-14th-edition-rjozm8zxcmm</loc>
    <lastmod>2026-07-19</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/RjozM8zXCmM/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-30: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A sphere is fired downwards into a medium with an initial speed of 27 meters per second. It experiences a deceleration given by a equals negative 6t meters per second squared, where t is time in seconds. Determine the total distance traveled before the sphere stops. Kinematics with Variable Acceleration: Finding Distance Traveled This lesson applies integral calculus to a kinematics problem involving a time-dependent (variable) deceleration. Students integrate acceleration to find velocity, determine when the object stops, and integrate velocity to compute total distance traveled. In this lesson, you will learn to: • Integrate a variable acceleration function to obtain a velocity function using an initial condition • Determine the time at which an object&apos;s velocity reaches zero by solving the resulting equation • Verify that velocity does not change sign over the interval o</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/RjozM8zXCmM</video:player_loc>
      <video:publication_date>2026-07-19T23:39:54.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-31-engineering-mechanics-dynamics-hibbeler-14th-edition-1aaokbmyco4</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/1aAOKBmyco4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-31: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: The velocity of a particle traveling along a straight line is v equals v-naught minus k-s, where k is a positive constant. If s equals zero when t equals zero, determine the position s as a function of time and the acceleration a as a function of time. Kinematics via Separable ODEs: Position and Acceleration from a Velocity Law This lesson applies first-order separable ordinary differential equations to a classical kinematics problem in which a particle&apos;s velocity depends linearly on its position. Students derive explicit expressions for position and acceleration as functions of time by separating variables, integrating, and applying an initial condition. In this lesson, you will learn to: • Set up and classify a first-order separable ODE from a given velocity–position relationship • Separate variables, integrate both sides, and determine the constant of integration from an</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/1aAOKBmyco4</video:player_loc>
      <video:publication_date>2026-07-20T12:05:14.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-32-engineering-mechanics-dynamics-hibbeler-14th-edition-qpwhyx32zr8</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/QpWhYX32ZR8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-32: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: Ball A is thrown vertically upward with an initial velocity v-naught. Ball B is thrown upward from the same point with the same initial velocity v-naught, but t seconds later. Given that t is less than 2 v-naught over g, determine the elapsed time measured from the instant ball A is thrown until the two balls pass each other. Also find the velocity of each ball at the instant they pass. Two Balls Thrown Upward at Different Times: Finding When They Meet This lesson applies rectilinear kinematics under constant gravitational acceleration to two balls launched vertically upward from the same point but at different times. Students derive the elapsed time at which the balls pass each other and find each ball&apos;s velocity at that instant, uncovering an elegant symmetry in the results. In this lesson, you will learn to: • Write position and velocity equations for objects undergoing c</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/QpWhYX32ZR8</video:player_loc>
      <video:publication_date>2026-07-20T12:05:54.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-33-engineering-mechanics-dynamics-hibbeler-14th-edition-2e0hoiulsdk</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/2E0HoIUlsdk/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-33: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: As a body is projected to a high altitude above the Earth&apos;s surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, the acceleration is given by: a equals negative g-naught times R squared over the quantity R plus y, all squared. Here, g-naught is the constant gravitational acceleration at sea level, R is the radius of the Earth, and the positive direction is measured upward. Given g-naught equals 9.81 meters per second squared and R equals 6356 kilometers, determine the minimum initial velocity — the escape velocity — at which a projectile should be shot vertically from the Earth&apos;s surface so that it does not fall back to the Earth. The required condition is that v equals zero as y approaches infinity. Escape Velocity via Separation of Variables This lesson derives the escape velocity of a projec</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/2E0HoIUlsdk</video:player_loc>
      <video:publication_date>2026-07-20T12:05:19.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-34-engineering-mechanics-dynamics-hibbeler-14th-edition-swvmvobanua</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/SwvmVoBanuA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-34: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: Accounting for the variation of gravitational acceleration a with respect to altitude y, derive an equation that relates the velocity of a freely falling particle to its altitude. Assume the particle is released from rest at altitude y-naught above Earth&apos;s surface. With what velocity does the particle strike Earth if it is released from rest at y-naught equals 500 kilometers? Use g equals 9.81 meters per second squared and R equals 6,371 kilometers. Velocity–Altitude Equation for Freefall with Variable Gravity This lesson derives a velocity–altitude relationship for a freely falling particle using Newton&apos;s inverse-square law of gravitation and the kinematic chain rule. The derivation integrates a separable ODE, producing a closed-form equation that is then evaluated numerically for a particle released from rest at an altitude of 500 km above Earth&apos;s surface. In this lesson,</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/SwvmVoBanuA</video:player_loc>
      <video:publication_date>2026-07-20T12:05:24.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-35-engineering-mechanics-dynamics-hibbeler-14th-edition-rb5tfq-tz3o</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/rb5Tfq_Tz3o/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-35: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A freight train starts from rest and travels with a constant acceleration of 0.5 feet per second squared. After a time t-prime it maintains a constant speed, so that when t equals 160 seconds it has traveled 2000 feet. Determine the time t-prime. Two-Phase Motion: Acceleration Then Constant Speed This lesson solves a two-phase kinematics problem in which an object accelerates uniformly from rest and then travels at constant speed. Students set up distance equations for each phase, combine them into a quadratic, and select the physically valid solution. In this lesson, you will learn to: • Break a multi-phase motion problem into acceleration and constant-speed segments and write a distance equation for each • Combine phase distances into a single quadratic equation and solve using the quadratic formula • Identify and reject physically invalid roots based on problem constrain</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/rb5Tfq_Tz3o</video:player_loc>
      <video:publication_date>2026-07-20T12:05:29.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-36-engineering-mechanics-dynamics-hibbeler-14th-edition-u6pguqyrsqc</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/u6pgUQYRsQc/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-36: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A particle&apos;s position-time graph has been experimentally determined. From the data, construct the velocity-time and acceleration-time graphs for the motion over the interval 0 to 40 seconds. For 0 to 30 seconds, the position follows the equation s equals 0.4 t squared meters. For t greater than or equal to 30 seconds, the position-time graph becomes a straight line. Constructing v–t and a–t Graphs from a Given s–t Graph This lesson demonstrates how to derive velocity–time and acceleration–time graphs from a piecewise position–time function using differentiation. The position function consists of a parabolic segment followed by a linear segment, requiring analysis of each region separately. Students apply the power rule and continuity conditions to fully characterize the motion. In this lesson, you will learn to: • Differentiate a piecewise position function to obtain veloci</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/u6pgUQYRsQc</video:player_loc>
      <video:publication_date>2026-07-20T12:05:33.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-37-engineering-mechanics-dynamics-hibbeler-14th-edition-a-flzy8mirq</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/A_FlzY8mIRQ/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-37: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: Two rockets start from rest at the same elevation. Rocket A accelerates vertically at 20 meters per second squared for 12 seconds and then maintains a constant speed. Rocket B accelerates at 15 meters per second squared until reaching a constant speed of 150 meters per second. Construct the acceleration-time, velocity-time, and position-time graphs for each rocket until t equals 20 seconds. What is the distance between the rockets when t equals 20 seconds? Piecewise Rocket Motion: a–t, v–t, and s–t Graphs This lesson analyzes two rockets undergoing piecewise rectilinear motion, each with a constant-acceleration phase followed by a constant-velocity phase. Students construct acceleration, velocity, and position graphs over a 20-second window and calculate the separation between the two rockets at t = 20 s. In this lesson, you will learn to: • Apply kinematic equations to eac</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/A_FlzY8mIRQ</video:player_loc>
      <video:publication_date>2026-07-20T12:05:37.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-38-engineering-mechanics-dynamics-hibbeler-14th-edition-mj-b1bf6sg8</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/MJ-B1BF6sG8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-38: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A particle starts from s equals 0 and travels along a straight line with a velocity v equals t squared minus 4t plus 3, measured in meters per second, where t is in seconds. Construct the v–t and a–t graphs for the time interval 0 less than or equal to t less than or equal to 4 seconds. Constructing v–t and a–t Graphs from a Velocity Function This lesson covers how to analyze a quadratic velocity function to construct velocity–time and acceleration–time graphs. Students differentiate the velocity function to find acceleration and identify key features such as zeros, the vertex, and endpoint values to accurately sketch each graph. In this lesson, you will learn to: • Find the zeros and vertex of a quadratic velocity function to identify when a particle is at rest and when velocity is at a minimum or maximum. • Differentiate a velocity function with respect to time to obtain t</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/MJ-B1BF6sG8</video:player_loc>
      <video:publication_date>2026-07-20T12:05:41.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-39-engineering-mechanics-dynamics-hibbeler-14th-edition-cy9jotdxuvq</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/CY9JoTDxuvQ/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-39: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A particle&apos;s position is defined by s equals 2 sine of pi t over 5, plus 4, in meters, where t is in seconds. Construct the s versus t, v versus t, and a versus t graphs for t ranging from 0 to 10 seconds. Graphing Position, Velocity, and Acceleration for Sinusoidal Motion This lesson derives and graphs the position, velocity, and acceleration functions for a particle undergoing simple harmonic motion defined by a sinusoidal position equation. Students apply differentiation using the chain rule to obtain velocity and acceleration, then evaluate each function at key time intervals to sketch all three kinematic graphs over one complete cycle. In this lesson, you will learn to: • Differentiate a sinusoidal position function using the chain rule to obtain velocity and acceleration functions. • Identify amplitude, angular frequency, period, and phase relationships among the s–t,</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/CY9JoTDxuvQ</video:player_loc>
      <video:publication_date>2026-07-20T12:05:46.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-40-engineering-mechanics-dynamics-hibbeler-14th-edition-eolu-r-xqby</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/EoLU-R_XqBY/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-40: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: An airplane starts from rest, travels 5000 feet down a runway, and after uniform acceleration takes off with a speed of 162 miles per hour. It then climbs in a straight line with a uniform acceleration of 3 feet per second squared until it reaches a constant speed of 220 miles per hour. Draw the s versus t, v versus t, and a versus t graphs that describe the motion. This lesson analyzes a three-phase rectilinear motion problem involving an airplane accelerating from rest on a runway, continuing to accelerate during a straight climb, and finally cruising at constant speed. Students learn to apply kinematic equations to each phase and interpret the resulting s–t, v–t, and a–t graphs. In this lesson, you will learn to: • Convert speeds between miles per hour and feet per second using dimensional analysis • Apply the constant-acceleration kinematic equations to find acceleration</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/EoLU-R_XqBY</video:player_loc>
      <video:publication_date>2026-07-20T12:05:50.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-41-engineering-mechanics-dynamics-hibbeler-14th-edition-wlcsh2lk2oy</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/WlCSh2lK2OY/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-41: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: An elevator starts from rest at ground level. It can accelerate at 5 feet per second squared and then decelerate at 2 feet per second squared. Determine the shortest time it takes to reach a point 40 feet above the starting position. The elevator starts from rest and stops at the upper level. Draw the acceleration-time, velocity-time, and position-time graphs for the motion. Problem 12-41: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson solves a rectilinear kinematics problem in which a body starts from rest, accelerates at one constant rate, then decelerates at a different constant rate until it stops over a fixed distance. Students find the shortest total travel time by equating peak velocities and applying kinematic equations, then interpret the resulting a–t, v–t, and s–t graphs. In this lesson, you will learn to: • Apply constant-acceleration kinemat</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/WlCSh2lK2OY</video:player_loc>
      <video:publication_date>2026-07-20T22:24:15.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-42-engineering-mechanics-dynamics-hibbeler-14th-edition-znvcikpf6mi</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/znVCiKPf6MI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-42: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: The velocity of a car is plotted as a v–t graph. From t equals 0 to t equals 40 seconds, the velocity is constant at v equals 10 meters per second. From t equals 40 seconds to t equals 80 seconds, the velocity decreases linearly from 10 meters per second to 0 meters per second. Determine the total distance the car travels until it stops at t equals 80 seconds, and construct the a–t graph. Problem 12-42: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson uses a piecewise linear velocity–time graph to determine acceleration and total distance traveled. Students learn to compute acceleration as the slope of the v–t curve and distance as the area beneath it, then construct the corresponding a–t graph. In this lesson, you will learn to: • Interpret a piecewise linear v–t graph to identify phases of constant velocity and constant acceleration • Calculate accelerat</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/znVCiKPf6MI</video:player_loc>
      <video:publication_date>2026-07-20T22:24:07.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-43-engineering-mechanics-dynamics-hibbeler-14th-edition-ry-wmcpeffo</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Ry-wmCpeFFo/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-43: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A jet plane lands on a runway with an initial speed of 300 feet per second, equal to 91.44 meters per second, at time t equals zero, and initial position s equals zero. The acceleration versus time relationship is given in three pieces. From zero to 10 seconds, acceleration equals zero. From 10 to 20 seconds, acceleration equals negative 2 times the quantity t minus 10, in meters per second squared, ramping linearly from zero down to negative 20 meters per second squared. For times greater than 20 seconds, acceleration is constant at negative 3 meters per second squared. Determine the time t-prime at which the plane comes to a stop, and construct the velocity versus time and position versus time graphs. Problem 12-43: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies piecewise integration of a given acceleration–time graph to construct the correspond</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Ry-wmCpeFFo</video:player_loc>
      <video:publication_date>2026-07-20T22:23:57.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-44-engineering-mechanics-dynamics-hibbeler-14th-edition-zewmg6si5t8</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/ZewMg6sI5T8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-44: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A particle travels between two plates spaced 200 millimeters apart. Its velocity-time graph has a symmetric triangular shape: the particle accelerates from rest at a constant 4 meters per second squared to a peak velocity v-max, then decelerates at 4 meters per second squared back to rest. Determine v-max and the total travel time t-prime. Also derive and describe the s-t graph, given that s equals 100 millimeters when t equals t-prime divided by 2. Problem 12-44: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson analyzes rectilinear particle motion defined by a symmetric triangular v–t graph, where equal constant acceleration and deceleration phases govern the motion. Students determine peak velocity and total travel time from a known displacement, then construct the corresponding s–t graph from piecewise kinematic equations. In this lesson, you will learn</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/ZewMg6sI5T8</video:player_loc>
      <video:publication_date>2026-07-20T22:24:03.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-45-engineering-mechanics-dynamics-hibbeler-14th-edition-rx-f03v44do</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/Rx-F03V44Do/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-45: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A particle moves through an electric field between two plates. Its velocity-time graph is a symmetric triangle, where t-prime equals 0.2 seconds and v-max equals 10 meters per second. Velocity increases linearly from zero to v-max over the first half of the time interval — that is, from 0 to t-prime over 2 — then decreases linearly from v-max back to zero over the second half, from t-prime over 2 to t-prime. At t equals t-prime over 2, which is 0.1 seconds, the particle is at position s equals 0.5 meters. Draw the s-versus-t and a-versus-t graphs for the particle. Rectilinear Kinematics: Constructing s–t and a–t Graphs from a Triangular v–t Graph This lesson uses a symmetric triangular velocity–time profile to derive both the position–time and acceleration–time graphs for a particle undergoing two phases of constant acceleration. Students practice integrating piecewise-linear</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/Rx-F03V44Do</video:player_loc>
      <video:publication_date>2026-07-20T22:23:54.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-46-engineering-mechanics-dynamics-hibbeler-14th-edition-1w17sn1e6bg</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/1W17Sn1e6bg/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-46: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A rocket moves along a straight track starting from rest — that is, v equals 0 at s equals 0. Its acceleration as a function of position is given in two pieces: a equals 5 feet per second squared for s between 0 and 100 feet, and a equals 5 plus 6 times the quantity root-s minus 10, raised to the 5-thirds power, in feet per second squared, for s greater than 100 feet. Determine the speed v when s equals 75 feet and when s equals 125 feet. Use Simpson&apos;s rule with n equals 100 intervals to evaluate the integral at s equals 125 feet. Problem 12-46: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies the kinematic identity a ds = v dv to find the speed of a rocket moving along a straight track, given a piecewise acceleration–position (a–s) function. Exact integration handles the constant-acceleration region, while Simpson&apos;s rule with 100 subintervals evalu</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/1W17Sn1e6bg</video:player_loc>
      <video:publication_date>2026-07-20T22:23:48.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-47-engineering-mechanics-dynamics-hibbeler-14th-edition-wbicbi4ng</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/__WBiCbI4Ng/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-47: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A two-stage rocket is fired vertically from rest at position zero. Its acceleration is described by a piecewise function. For times from 0 up to 30 seconds, the acceleration increases linearly from 0 to 12 meters per second squared, given by the expression two-fifths times t. For times from 30 to 60 seconds, the acceleration is constant at 24 meters per second squared. Plot the velocity versus time graph and the position versus time graph describing the rocket&apos;s motion for times from 0 to 60 seconds. Problem 12-47: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson solves a rectilinear kinematics problem in which a rocket launches vertically from rest under a piecewise-defined acceleration: linearly increasing during the first phase and constant during the second. Students integrate the acceleration to build the velocity–time graph and then integrate again to</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/__WBiCbI4Ng</video:player_loc>
      <video:publication_date>2026-07-20T22:23:44.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-48-engineering-mechanics-dynamics-hibbeler-14th-edition-op-yhrvnw4k</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/oP_yHRvnw4k/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-48: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A race car starts from rest and travels along a straight road until it reaches a speed of 26 meters per second in 8 seconds. The v–t graph consists of three segments: v equals 3.5 times t, for t from 0 up to 4 seconds; v equals 14 meters per second, constant, for t from 4 seconds to 5 seconds; and v equals 4t minus 6, for t from 5 seconds to 8 seconds. Draw the a–t graph and determine the maximum acceleration of the car. Problem 12-48: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson shows how to derive the acceleration–time graph from a piecewise-linear velocity–time graph by differentiating each linear segment. Students practice applying the definition a = dv/dt to obtain constant acceleration values on each interval and identify the maximum acceleration. In this lesson, you will learn to: • Apply the definition a = dv/dt to find acceleration from a piec</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/oP_yHRvnw4k</video:player_loc>
      <video:publication_date>2026-07-20T22:23:38.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-49-engineering-mechanics-dynamics-hibbeler-14th-edition-1q-9-2xmrww</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/1Q-9_2xMrww/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-49: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A jet car is originally traveling at a velocity of 10 meters per second when it is subjected to a piecewise acceleration. The acceleration equals 6 meters per second squared for times from 0 up to 15 seconds, and then equals negative 4 meters per second squared for all times at or after 15 seconds. At time zero, the position is zero. Determine, first, the car&apos;s maximum velocity, and second, the time t-prime when the car stops. Problem 12-49: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson analyzes the straight-line motion of a vehicle subjected to two different constant accelerations applied in sequence. Students use direct integration of the equation of motion to build piecewise velocity functions, identify the maximum velocity, and determine when the vehicle comes to rest. In this lesson, you will learn to: • Integrate piecewise constant acceleration fu</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/1Q-9_2xMrww</video:player_loc>
      <video:publication_date>2026-07-20T22:23:33.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-50-engineering-mechanics-dynamics-hibbeler-14th-edition-vnryhvem54</loc>
    <lastmod>2026-07-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/-VnRyhVEm54/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-50: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A car starts from rest at s equals 0 and is subjected to an acceleration defined by the following a-s graph. For s from 0 to 300 feet, acceleration equals 12 feet per second squared. For s from 300 to 450 feet, acceleration equals negative 0.04 times s plus 24 feet per second squared, which decreases linearly from 12 feet per second squared at s equals 300 feet down to 6 feet per second squared at s equals 450 feet. Draw the v-s graph and determine the time needed to travel 200 feet. Problem 12-50: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson develops velocity–position (v–s) relationships directly from a piecewise acceleration–position (a–s) graph using the kinematic identity a ds = v dv. Students construct the v–s graph analytically and then integrate v = ds/dt to determine the time required to reach a specified position. In this lesson, you will lear</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/-VnRyhVEm54</video:player_loc>
      <video:publication_date>2026-07-20T22:23:29.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-51-engineering-mechanics-dynamics-hibbeler-14th-edition-ev7uv7yjgfq</loc>
    <lastmod>2026-07-21</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/eV7uV7yjGfQ/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-51: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A velocity-time graph for a train consists of three segments. From 0 to 60 seconds, velocity increases linearly from 0 to 6 meters per second. From 60 seconds to 120 seconds, velocity is constant at 6 meters per second. From 120 seconds to 180 seconds, velocity increases linearly from 6 to 10 meters per second. Given that position s equals 0 when t equals 0, construct the position-time and acceleration-time graphs for t between 0 and 180 seconds. Problem 12-51: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson uses a piecewise linear velocity–time graph for a train to derive the corresponding position–time and acceleration–time graphs. Students integrate each velocity segment to obtain position functions and differentiate to obtain constant acceleration values, producing a smooth piecewise-parabolic s–t curve and a step-function a–t graph. In this lesson, y</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/eV7uV7yjGfQ</video:player_loc>
      <video:publication_date>2026-07-21T00:34:48.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-52-engineering-mechanics-dynamics-hibbeler-14th-edition-9o3evvr3u-i</loc>
    <lastmod>2026-07-21</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/9o3eVvr3u-I/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-52: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A motorcycle starts from rest at s equals 0 and travels along a straight road. Its speed is given by a piecewise v–t graph with three segments. From t equals 0 to 4 seconds, velocity equals 1.25 times t meters per second, increasing linearly from rest to 5 meters per second. From t equals 4 to 10 seconds, velocity is constant at 5 meters per second. From t equals 10 to 15 seconds, velocity equals negative t plus 15 meters per second, decreasing linearly back to zero. Determine the total distance the motorcycle travels until it stops at t equals 15 seconds, and describe the shapes of the a–t and s–t graphs. Problem 12-52: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson analyzes the motion of a vehicle given a piecewise linear velocity–time graph with three segments: uniform acceleration, constant speed, and uniform deceleration. Students compute acceleratio</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/9o3eVvr3u-I</video:player_loc>
      <video:publication_date>2026-07-21T00:34:52.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-53-engineering-mechanics-dynamics-hibbeler-14th-edition-qkln-tn7kxy</loc>
    <lastmod>2026-07-21</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/QkLN_tN7KXY/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-53: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A motorcycle starts from rest at s equals 0 and travels along a straight road with the speed shown by a velocity-time graph. The graph has three segments: from 0 to 4 seconds, v equals 1.25 t; from 4 to 10 seconds, v equals 5 meters per second; and from 10 to 15 seconds, v equals negative t plus 15. Determine the motorcycle&apos;s acceleration and position when t equals 8 seconds and when t equals 12 seconds. Problem 12-53: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson uses a piecewise linear velocity–time graph to determine a vehicle&apos;s acceleration and position at specific instants. Students apply the principles that acceleration equals the slope of the v–t graph and displacement equals the area under the v–t graph. In this lesson, you will learn to: • Interpret a piecewise linear v–t graph and identify the velocity function for each segment • Calculate acc</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/QkLN_tN7KXY</video:player_loc>
      <video:publication_date>2026-07-21T00:34:57.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-54-engineering-mechanics-dynamics-hibbeler-14th-edition-3fxgbj4fhau</loc>
    <lastmod>2026-07-21</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/3FxGBj4fHAU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-54: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: The velocity-time graph for a car moving along a straight road is defined as follows: velocity equals 0.6 t squared, for t between 0 and 5 seconds; and a straight line dropping from v equals 15 meters per second at t equals 5 seconds, down to v equals 0 at t equals 15 seconds. At time t equals 0, the position s equals 0. Part one: determine s as a function of t and a as a function of t for each interval. Part two: find the average speed and total distance traveled over the full 15-second interval. Problem 12-54: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson derives position and acceleration functions from a piecewise velocity–time graph for a car moving along a straight road. Students integrate and differentiate the velocity functions over two intervals, then compute total distance traveled and average speed over a 15-second window. In this lesson, you</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/3FxGBj4fHAU</video:player_loc>
      <video:publication_date>2026-07-21T00:35:02.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-55-engineering-mechanics-dynamics-hibbeler-14th-edition-c8wvn8u1ygs</loc>
    <lastmod>2026-07-21</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/C8WVn8U1YGs/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-55: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: An airplane lands on a straight runway, originally traveling at 110 feet per second when position s equals zero. It is subjected to the following piecewise constant decelerations: acceleration equals 0 feet per second squared for time between 0 and 5 seconds; acceleration equals negative 3 feet per second squared for time between 5 and 15 seconds; acceleration equals negative 8 feet per second squared for time between 15 and 20 seconds; and acceleration equals negative 3 feet per second squared for time from 20 seconds to t-prime. Determine the time t-prime needed to stop the plane, and construct the s-t graph for the motion. Rectilinear Kinematics with Piecewise Constant Acceleration This lesson solves a piecewise constant-acceleration problem for an airplane decelerating along a straight runway. Students apply the standard kinematic equations segment by segment to find the</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/C8WVn8U1YGs</video:player_loc>
      <video:publication_date>2026-07-21T00:35:06.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-56-engineering-mechanics-dynamics-hibbeler-14th-edition-d5ze-zkux3k</loc>
    <lastmod>2026-07-21</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/D5zE_Zkux3k/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-56: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A boat starts from rest at a position of zero feet and travels in a straight line. Its acceleration varies with position according to a piecewise linear graph. From zero to 100 feet, acceleration decreases linearly from 8 feet per second squared down to 6 feet per second squared. From 100 feet to 150 feet, acceleration continues to decrease linearly from 6 feet per second squared down to zero. Determine the boat&apos;s speed when it has traveled 50 feet, 100 feet, and 150 feet. Problem 12-56: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson demonstrates how to determine a moving object&apos;s speed at specified positions using an acceleration-displacement (a-s) graph. Students apply the kinematic relation v dv = a ds, integrate piecewise linear acceleration functions, and compute velocities from the area under the a-s curve. In this lesson, you will learn to: • Appl</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/D5zE_Zkux3k</video:player_loc>
      <video:publication_date>2026-07-21T10:26:32.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-57-engineering-mechanics-dynamics-hibbeler-14th-edition-d2o1x9hoil8</loc>
    <lastmod>2026-07-21</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/D2o1X9hoIl8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-57: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A boat starts from rest at s equals 0 and travels in a straight line. The acceleration varies as follows: a decreases linearly from 8 feet per second squared to 6 feet per second squared as s goes from 0 to 100 feet, then decreases linearly from 6 feet per second squared to 0 feet per second squared as s goes from 100 feet to 150 feet. Construct the v–s graph. Problem 12-57: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson demonstrates how to build a velocity–position (v–s) graph for a particle undergoing rectilinear motion when acceleration varies linearly with position in two successive segments. The method centers on integrating the kinematic identity v dv = a ds piecewise and verifying continuity at the segment boundary. In this lesson, you will learn to: • Apply the kinematic identity v dv = a ds to relate velocity and position under variable accelera</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/D2o1X9hoIl8</video:player_loc>
      <video:publication_date>2026-07-21T10:26:36.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-58-engineering-mechanics-dynamics-hibbeler-14th-edition-6k62drm3gnu</loc>
    <lastmod>2026-07-21</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/6K62DrM3gNU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-58: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A two-stage rocket is fired vertically from rest. In Phase 1, from time zero to 15 seconds, the acceleration increases linearly from zero to 15 meters per second squared, so a of t equals t. In Phase 2, from 15 to 40 seconds, the acceleration is constant at 20 meters per second squared. The initial velocity and initial position are both zero. Derive and plot the velocity-time and position-time graphs for time from zero to 40 seconds. Problem 12-58: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson analyzes the rectilinear motion of a two-stage rocket using integration of a piecewise acceleration function. Students derive velocity and position as functions of time for both a linearly increasing acceleration phase and a constant acceleration phase, then identify the shape of the resulting v–t and s–t graphs. In this lesson, you will learn to: • Derive velocit</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/6K62DrM3gNU</video:player_loc>
      <video:publication_date>2026-07-21T10:37:09.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-59-engineering-mechanics-dynamics-hibbeler-14th-edition-0dj-axad0b4</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/0dJ-axaD0b4/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-59: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: The speed of a train during the first minute has been recorded at four moments. At time 0 seconds, the velocity is 0 meters per second. At 20 seconds, the velocity is 16 meters per second. At 40 seconds, the velocity is 21 meters per second. At 60 seconds, the velocity is 24 meters per second. Approximating the v-t curve as straight-line segments between the given points, determine the total distance traveled. Finding Distance from a v–t Graph Using the Trapezoidal Rule This lesson demonstrates how to determine the total distance traveled by an object when velocity is recorded at discrete time intervals. Students learn to approximate the v–t curve with straight-line segments and calculate the area under the graph using the trapezoidal rule for each interval. In this lesson, you will learn to: • Interpret a velocity–time graph constructed from discrete data points connected b</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/0dJ-axaD0b4</video:player_loc>
      <video:publication_date>2026-07-22T10:03:42.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-60-engineering-mechanics-dynamics-hibbeler-14th-edition-jiwopeedsac</loc>
    <lastmod>2026-07-21</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/JiwOPEeDsAc/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-60: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: A package is dropped from an elevator at a height of 100 feet above the ground. The elevator maintains a constant upward speed of 4 feet per second. The package is released with the same upward speed as the elevator. Assume gravitational acceleration is 32 feet per second squared. First, determine the height of the elevator from the ground at the instant the package hits the ground. Second, draw the velocity-time curve for the package during its motion. Problem 12-60: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson applies one-dimensional kinematics to a package released from an upward-moving elevator. Students set up position and velocity equations under constant gravitational acceleration, solve a quadratic for impact time, and track a second object moving at constant velocity to find its position at the same instant. In this lesson, you will learn to:</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/JiwOPEeDsAc</video:player_loc>
      <video:publication_date>2026-07-21T10:26:41.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-12-61-engineering-mechanics-dynamics-hibbeler-14th-edition-kvprywzxrau</loc>
    <lastmod>2026-07-22</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/kVpryWzxRAU/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 12-61: Engineering Mechanics- Dynamics (Hibbeler 14th Edition)</video:title>
      <video:description>Question: Two cars start from rest side by side and travel along a straight road. Car A accelerates at 4 meters per second squared for 10 seconds and then maintains a constant speed. Car B accelerates at 5 meters per second squared until reaching a constant speed of 25 meters per second and then maintains this speed. Construct the acceleration–time, velocity–time, and position–time graphs for each car until t equals 15 seconds. What is the distance between the two cars when t equals 15 seconds? Problem 12-61: Engineering Mechanics- Dynamics (Hibbeler 14th Edition) This lesson analyzes the rectilinear motion of two cars that start from rest and undergo constant acceleration before reaching constant speeds. Students construct acceleration–time, velocity–time, and position–time graphs for each car over a 15-second interval and determine the separation distance between them at a specific i</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/kVpryWzxRAU</video:player_loc>
      <video:publication_date>2026-07-22T10:03:49.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-5-10-engineering-mechanics-statics-by-russell-c-hibbeler-1hl-kov0j4y</loc>
    <lastmod>2023-02-07</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/1Hl-kOv0J4Y/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-10 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>Determine the components of the support reactions at the fixed support A on the cantilevered beam. You can download the pdf solution file freely from doassignment.ca Don&apos;t hesitate to ask your questions in the comment.</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/1Hl-kOv0J4Y</video:player_loc>
      <video:publication_date>2023-02-07T05:32:57.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
    </video:video>
  </url>
  <url>
    <loc>https://doassignment.ca/videos/problem-5-11-engineering-mechanics-statics-by-russell-c-hibbeler-ptkx6jauia8</loc>
    <lastmod>2023-02-07</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/pTKX6Jauia8/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-11 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>Determine the reactions at the supports. In this video, I want to show you how to solve question 11 chapter 5 of Engineering Mechanics_ Statics by Russell C. Hibbeler. At point A, we have a roller joint, and at point B we have a revolute joint. To download the solution in the pdf format go to www.doassignment.ca</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/pTKX6Jauia8</video:player_loc>
      <video:publication_date>2023-02-07T05:33:03.000Z</video:publication_date>
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    <lastmod>2023-02-07</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/SSPVmGmVx24/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-12 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>5–12. Determine the horizontal and vertical components of reaction at the pin A and the reaction of the rocker B on the beam. SUBSCRIBE to my Channel for more problem Solutions! Kindly like, share, and comment, this will help to promote my channel! Equilibrium of a Rigid Body (2D Equilibrium) How to solve Equilibrium Problems | Doassignment Engineering Statics by Hibbeler 14th Edition</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/SSPVmGmVx24</video:player_loc>
      <video:publication_date>2023-02-07T17:17:53.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
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    <loc>https://doassignment.ca/videos/problem-5-13-engineering-mechanics-statics-by-russell-c-hibbeler-vuu-docr1ts</loc>
    <lastmod>2023-02-08</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/vuu_Docr1ts/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-13 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>5–13. Determine the reactions at the supports. SUBSCRIBE to my Channel for more problem Solutions! Kindly like, share, and comment, this will help to promote my channel! Equilibrium of a Rigid Body (2D Equilibrium) How to solve Equilibrium Problems | Doassignment Engineering Statics by Hibbeler 14th Edition</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/vuu_Docr1ts</video:player_loc>
      <video:publication_date>2023-02-08T06:14:42.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
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  <url>
    <loc>https://doassignment.ca/videos/problem-5-14-engineering-mechanics-statics-by-russell-c-hibbeler-kvl1lbwvh7o</loc>
    <lastmod>2023-02-17</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/KVl1lBWVH7o/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-14 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>5–13. Determine the reactions at the supports. SUBSCRIBE to my Channel for more problem Solutions! Kindly like, share, and comment, this will help to promote my channel! Equilibrium of a Rigid Body (2D Equilibrium) How to solve Equilibrium Problems | Doassignment Engineering Statics by Hibbeler 14th Edition</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/KVl1lBWVH7o</video:player_loc>
      <video:publication_date>2023-02-17T06:35:28.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
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  <url>
    <loc>https://doassignment.ca/videos/problem-5-15-engineering-mechanics-statics-by-russell-c-hibbeler-x-1z1ze11qw</loc>
    <lastmod>2023-02-17</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/X_1z1ZE11qw/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-15 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>5–15. Determine the reactions at the supports. SUBSCRIBE to my Channel for more problem Solutions! Kindly like, share, and comment, this will help to promote my channel! Equilibrium of a Rigid Body (2D Equilibrium) How to solve Equilibrium Problems | Doassignment Engineering Statics by Hibbeler 14th Edition</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/X_1z1ZE11qw</video:player_loc>
      <video:publication_date>2023-02-17T21:42:26.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
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  <url>
    <loc>https://doassignment.ca/videos/problem-5-16-engineering-mechanics-statics-by-russell-c-hibbeler-dct8ffv55gm</loc>
    <lastmod>2023-02-20</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/dCT8fFV55GM/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-16 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>5–16. Determine the tension in the cable and the horizontal and vertical components of reaction of the pin A. The pulley at D is frictionless and the cylinder weighs 80 lb. Please consider subscribing to my channel for more problem solutions. Your likes, shares, and comments are greatly appreciated and help to promote my channel. Thank you! Equilibrium of a Rigid Body (2D Equilibrium) How to solve Equilibrium Problems | Doassignment Engineering Statics by Hibbeler 14th Edition</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/dCT8fFV55GM</video:player_loc>
      <video:publication_date>2023-02-20T19:03:50.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
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  <url>
    <loc>https://doassignment.ca/videos/problem-5-17-engineering-mechanics-statics-by-russell-c-hibbeler-idzoqlvdyoi</loc>
    <lastmod>2023-03-15</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/idZOqlvdyoI/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-17 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>5–17. The man attempts to support the load of boards having a weight W and a center of gravity at G. If he is standing on a smooth floor, determine the smallest angle at which he can hold them up in the position shown. Neglect his weight. Please consider subscribing to my channel for more problem solutions. Your likes, shares, and comments are greatly appreciated and help to promote my channel. Thank you! Equilibrium of a Rigid Body (2D Equilibrium) How to solve Equilibrium Problems | Doassignment Engineering Statics by Hibbeler 14th Edition</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/idZOqlvdyoI</video:player_loc>
      <video:publication_date>2023-03-15T17:32:06.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
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  <url>
    <loc>https://doassignment.ca/videos/problem-5-18-engineering-mechanics-statics-by-russell-c-hibbeler-aq24j4nuas0</loc>
    <lastmod>2023-05-03</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/AQ24j4NUAs0/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-18 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>5–18. Determine the components of reaction at the supports A and B on the rod. In sigma fx=0 it should be written B-P=0, but that doesn&apos;t make any change in the answer. Please consider subscribing to my channel for more problem solutions. Your likes, shares, and comments are greatly appreciated and help to promote my channel. Thank you! Equilibrium of a Rigid Body (2D Equilibrium) How to solve Equilibrium Problems | Doassignment Engineering Statics by Hibbeler 14th Edition</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/AQ24j4NUAs0</video:player_loc>
      <video:publication_date>2023-05-03T15:46:55.000Z</video:publication_date>
      <video:family_friendly>yes</video:family_friendly>
      <video:live>no</video:live>
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  <url>
    <loc>https://doassignment.ca/videos/problem-5-19-engineering-mechanics-statics-by-russell-c-hibbeler-sjh0-nqtjra</loc>
    <lastmod>2023-05-04</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/SjH0-NqTjRA/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-19 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>5–19. The man has a weight W and stands at the center of the plank. If the planes at A and B are smooth, determine the tension in the cord in terms of W and theta. Please consider subscribing to my channel for more problem solutions. Your likes, shares, and comments are greatly appreciated and help to promote my channel. Thank you! Equilibrium of a Rigid Body (2D Equilibrium) How to solve Equilibrium Problems | Doassignment Engineering Statics by Hibbeler 14th Edition</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/SjH0-NqTjRA</video:player_loc>
      <video:publication_date>2023-05-04T05:31:56.000Z</video:publication_date>
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  <url>
    <loc>https://doassignment.ca/videos/problem-5-21-engineering-mechanics-statics-by-russell-c-hibbeler-esnidzf0meq</loc>
    <lastmod>2026-07-21</lastmod>
    <video:video>
      <video:thumbnail_loc>https://i.ytimg.com/vi/ESniDZF0mEQ/mqdefault.jpg</video:thumbnail_loc>
      <video:title>Problem 5-21 Engineering Mechanics_ Statics by Russell C. Hibbeler</video:title>
      <video:description>Question: A uniform rod A B has a mass of 40 kilograms. The rod is 3 meters long and is positioned at 60 degrees from the horizontal. End A slides along a smooth vertical wall using a smooth collar, and end B rests on a smooth horizontal floor with a cable attached at B running horizontally to the base of the wall at C. Determine the tension in cable B C. Problem 5-21 Engineering Mechanics_ Statics by Russell C. Hibbeler This lesson applies 2D rigid-body equilibrium to a uniform rod supported by a smooth collar on a vertical wall, a smooth floor, and a horizontal cable. Students use free-body diagrams and the three scalar equilibrium equations — including a strategic moment equation — to determine the unknown cable tension. In this lesson, you will learn to: • Draw a correct free-body diagram identifying all reaction forces on a rod with a smooth collar, smooth floor contact, and a ca</video:description>
      <video:player_loc allow_embed="yes">https://www.youtube.com/embed/ESniDZF0mEQ</video:player_loc>
      <video:publication_date>2026-07-21T08:41:39.000Z</video:publication_date>
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