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B2.1 · Use position, displacement, speed, velocity, and acceleration terminology

Learn to use position, displacement, speed, velocity, and acceleration terminology through clear examples and targeted practice.

Ontario Grade 11 Physics

Kinematics

A Grade 11 guide to the language used to describe how an object moves

Motion is easier to describe when we agree on what object we are tracking and where its position is measured from. In this lesson, the system is one moving object, such as a person or a cart. We will use a straight path and choose one direction as positive. A position tells where the object is relative to a reference point. A displacement describes the change in position. Speed describes how quickly distance is covered, while velocity also includes direction. Acceleration describes how velocity changes over time. These terms are related, but they do not mean the same thing.

What you will learn

1. Prerequisite bridge: reference points, signs, and quantities

A reference point is a chosen location used to describe where something is. Position is measured from this point. For example, if a doorway is the reference point, a location 4 m to its right can be written as +4 m when right is chosen as positive. A location 2 m to its left is then −2 m. The plus and minus signs show direction; they do not mean that one position is better or worse.
A scalar has magnitude, or size, but no direction. Time and speed are scalars. A vector has both magnitude and direction. Position, displacement, velocity, and acceleration are vectors in this lesson. A signed value on a straight path is a simple way to show a vector's direction: positive means the chosen positive direction, and negative means the opposite direction.
Use SI units, the standard metric units used in science. Position and displacement are measured in metres (m), time in seconds (s), speed and velocity in metres per second (m/s), and acceleration in metres per second squared (m/s²). Before describing motion, state the object, reference point, and positive direction.

2. Position and displacement

Position, written as xx, tells an object's location relative to the reference point. It includes direction when the object is on a straight path. For example, x=−3.0 mx=−3.0\,\mathrm{m} means the object is 3.0 m in the direction opposite to the chosen positive direction.
Displacement, written as Δx\Delta x, is the change in position from an initial position to a final position. The symbol Δ\Delta means “change in.” Subtract the initial position from the final position. The result is a vector: its sign gives the direction of the change. Displacement does not describe every part of the route; it compares only the start and finish positions.
Imagine a person walking 5 m to the right and then 2 m to the left. If right is positive, the person's final position is 3 m to the right of the start. The displacement is therefore +3 m, even though the person walked a total distance of 7 m. Distance is the length of the route travelled and is a scalar. Keeping distance and displacement separate prevents a common mix-up.
Δx=xf−xi\Delta x=x_{\mathrm{f}}-x_{\mathrm{i}}

3. Speed and velocity

Speed tells how much distance is covered per unit of time. It is a scalar, so it has no direction. Average speed is found by dividing the total distance travelled by the elapsed time. Elapsed time is the time at the end minus the time at the start.
Velocity tells how quickly position changes and in what direction. Average velocity is displacement divided by elapsed time. Its sign gives the direction of the displacement. On a straight path, a positive velocity means motion in the chosen positive direction; a negative velocity means motion in the opposite direction. A zero average velocity means the final and initial positions are the same. It does not prove that the object stayed still throughout the interval.
Speed and the magnitude of velocity can have the same numerical value for motion in one direction. They are still different terms: speed has no direction, while velocity does. For example, 3.0 m/s is a speed. A velocity could be +3.0 m/s, with the sign tied to the chosen direction.
average speed=total distanceelapsed time,average velocity=ΔxΔt\text{average speed}=\frac{\text{total distance}}{\text{elapsed time}},\quad \text{average velocity}=\frac{\Delta x}{\Delta t}

4. Acceleration: change in velocity

Acceleration describes how velocity changes over time. Average acceleration is the change in velocity divided by the elapsed time. Since velocity has direction, acceleration also has direction. Its SI unit is m/s², read as metres per second squared. This unit means that velocity changes by a certain number of metres per second during each second.
A positive acceleration does not always mean an object is speeding up. It means the velocity is changing in the positive direction. If velocity and acceleration point in the same direction, the object's speed increases. If they point in opposite directions, its speed decreases. For instance, a velocity of −6 m/s changing toward zero has a positive change in velocity and positive acceleration, even though the object is slowing down.
Use signed velocities consistently. Find the change in velocity by subtracting the initial velocity from the final velocity, then divide by the time interval. Check the sign of the result against the chosen positive direction. Acceleration can be positive, negative, or zero; zero average acceleration means there was no net change in velocity over that interval.
aavg=vf−viΔta_{\mathrm{avg}}=\frac{v_{\mathrm{f}}-v_{\mathrm{i}}}{\Delta t}

Worked example

1. Finding displacement from two positions

A person is at position −2.0 m, then walks to position +5.0 m. The reference point is the starting marker, and right is positive. Find the person's displacement.
  1. Set the direction and identify values
    The system is the person. Right is positive. The initial position is −2.0 m and the final position is +5.0 m. The unknown is displacement.
    xi=−2.0 m,xf=+5.0 mx_{\mathrm{i}}=−2.0\,\mathrm{m},\quad x_{\mathrm{f}}=+5.0\,\mathrm{m}
  2. Use the displacement relationship
    Displacement is final position minus initial position. Subtracting a negative position adds its magnitude, so the result is positive.
    Δx=xf−xi=(+5.0 m)−(−2.0 m)=+7.0 m\Delta x=x_{\mathrm{f}}-x_{\mathrm{i}}=(+5.0\,\mathrm{m})-(−2.0\,\mathrm{m})=+7.0\,\mathrm{m}
Answer: The displacement is +7.0 m, or 7.0 m to the right.
Check: The unit is metres, as required for displacement. The positive sign agrees with the person's net change toward the right. A change from −2.0 m to +5.0 m spans 7.0 m, so the size is reasonable.

Worked example

2. Comparing average speed and average velocity

A cart moves 12.0 m east, then 4.0 m west in 8.0 s. Let east be positive. Find its average speed and average velocity.
  1. Describe the system and direction
    The system is the cart, and east is positive. Its total distance is the sum of both parts of the route. Its displacement is the signed change from start to finish.
    dtotal=12.0 m+4.0 m=16.0 m,Δx=+12.0 m−4.0 m=+8.0 md_{\mathrm{total}}=12.0\,\mathrm{m}+4.0\,\mathrm{m}=16.0\,\mathrm{m},\quad \Delta x=+12.0\,\mathrm{m}−4.0\,\mathrm{m}=+8.0\,\mathrm{m}
  2. Calculate the average speed
    Average speed uses total distance divided by elapsed time. It is positive because speed has no direction.
    average speed=16.0 m8.0 s=2.0 m/s\text{average speed}=\frac{16.0\,\mathrm{m}}{8.0\,\mathrm{s}}=2.0\,\mathrm{m/s}
  3. Calculate the average velocity
    Average velocity uses displacement divided by elapsed time. The positive result indicates east.
    average velocity=+8.0 m8.0 s=+1.0 m/s\text{average velocity}=\frac{+8.0\,\mathrm{m}}{8.0\,\mathrm{s}}=+1.0\,\mathrm{m/s}
Answer: The average speed is 2.0 m/s. The average velocity is +1.0 m/s, or 1.0 m/s east.
Check: Distance divided by time and displacement divided by time both have units of m/s. The cart travelled 16.0 m but ended only 8.0 m east of its start, so its average speed is greater than the magnitude of its average velocity, as expected.

Worked example

3. Finding acceleration from signed velocities

A cyclist travels west at 2.0 m/s, then travels east at 4.0 m/s, 3.0 s later. Let east be positive. Find the average acceleration over this interval.
  1. Set the system and record signed velocities
    The system is the cyclist, and east is positive. Westward initial velocity is negative; eastward final velocity is positive. The unknown is average acceleration.
    vi=−2.0 m/s,vf=+4.0 m/s,Δt=3.0 sv_{\mathrm{i}}=−2.0\,\mathrm{m/s},\quad v_{\mathrm{f}}=+4.0\,\mathrm{m/s},\quad \Delta t=3.0\,\mathrm{s}
  2. Find the change in velocity per time
    Subtract the initial velocity from the final velocity, then divide by the elapsed time. The positive result means the velocity changed toward the east.
    aavg=(+4.0 m/s)−(−2.0 m/s)3.0 s=+2.0 m/s2a_{\mathrm{avg}}=\frac{(+4.0\,\mathrm{m/s})−(−2.0\,\mathrm{m/s})}{3.0\,\mathrm{s}}=+2.0\,\mathrm{m/s^2}
Answer: The average acceleration is +2.0 m/s², or 2.0 m/s² east.
Check: The velocity change is 6.0 m/s over 3.0 s, so the change is 2.0 m/s each second. The units reduce to m/s², and the positive direction agrees with the change from westward to eastward velocity.

Common mistakes and how to avoid them

Treating position and displacement as the same quantity.
Correction: Position is a location relative to a reference point. Displacement is the difference between final and initial positions.
Using total distance to calculate average velocity.
Correction: Use displacement for average velocity. Use total distance for average speed.
Thinking a negative velocity means the object is slowing down.
Correction: A negative velocity indicates motion opposite the chosen positive direction. To tell whether speed is changing, compare the directions of velocity and acceleration.
Thinking positive acceleration always means speeding up.
Correction: Positive acceleration indicates a change in velocity toward the positive direction. An object moving in the negative direction may slow down while its acceleration is positive.

Lesson summary

Check your understanding

Question 1

A student starts at +3 m and ends at −1 m. If right is positive, what is the displacement?
  1. −4 m
  2. +4 m
  3. −1 m
  4. +2 m
Show answer and explanation
−4 m
Displacement is final position minus initial position: −1 m − (+3 m) = −4 m. The negative sign means left.

Question 2

A runner covers 30 m in 6.0 s. What is the runner's average speed?
  1. 5.0 m/s
  2. 0.20 m/s
  3. 36 m/s
  4. −5.0 m/s
Show answer and explanation
5.0 m/s
Average speed is total distance divided by time: 30 m ÷ 6.0 s = 5.0 m/s. Speed is a scalar, so it is not negative.

Question 3

East is positive. An object's velocity changes from +7 m/s to +3 m/s in 2.0 s. What is its average acceleration?
  1. −2.0 m/s²
  2. +2.0 m/s²
  3. −5.0 m/s²
  4. +5.0 m/s²
Show answer and explanation
−2.0 m/s²
Average acceleration is (final velocity − initial velocity) ÷ time: (+3 − +7) m/s ÷ 2.0 s = −2.0 m/s². The negative result indicates a change toward the west.

Key terms

Reference point
A chosen location used to describe an object's position.
Scalar
A quantity with magnitude but no direction.
Vector
A quantity with both magnitude and direction.
Displacement
The change in position from an initial position to a final position.
Average speed
Total distance travelled divided by elapsed time.
Average velocity
Displacement divided by elapsed time.
Acceleration
The change in velocity divided by the time over which that change occurs.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation B2.1. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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