B2.7 Solve uniform and accelerated motion problems graphically and algebraically

Ontario Grade 11 Physics

Kinematics

Reading motion graphs and choosing algebraic relationships

Motion problems describe how an object’s position or velocity changes over time. A scalar has size only, such as time or distance. A vector has size and direction, such as displacement, velocity, and acceleration. Before solving, name the moving object as the system, choose a positive direction, and keep that choice throughout. In this lesson, right or east is positive unless a problem says otherwise. Motion graphs show how quantities change; algebra uses relationships among the same quantities. Both methods should lead to answers that agree.

What you will learn

Distinguish scalar quantities from motion vectors and state a positive direction.
Read position–time and velocity–time graphs to describe motion.
Solve uniform-motion and constant-acceleration problems graphically and algebraically.
Check units, direction, significant figures, and whether an answer makes sense.

1. Prerequisite bridge: quantities, signs, and units

Position tells where an object is relative to a chosen origin. Displacement is the change in position, so it includes direction. Distance is the total path length and is not negative. For example, a change in position of

-4.0\ \mathrm{m}

means a displacement of

4.0\ \mathrm{m}

in the negative direction.

Time is measured in seconds, position and displacement in metres, velocity in metres per second, and acceleration in metres per second squared. Velocity describes the rate and direction of position change. Acceleration describes the rate and direction of velocity change. A negative value means the vector points opposite to the chosen positive direction; it does not automatically mean the object is slowing down.

For uniform motion, velocity is constant. For uniformly accelerated motion, acceleration is constant. The symbol

\Delta

means “change in.” Subscripts

i

and

f

mean initial and final. Algebra skills such as rearranging an equation are enough to use the relationships in this lesson.

\Delta x=x_f-x_i

Choose and state the system, origin, and positive direction before calculating.
A negative vector value describes direction relative to the chosen axis.
Use SI units: seconds, metres, metres per second, and metres per second squared.

2. What motion graphs show

A position–time graph has time on the horizontal axis and position on the vertical axis. Its slope gives velocity: a steeper line means a greater velocity magnitude. A straight line has constant slope and represents uniform motion. A horizontal line means the position is unchanged, so the object is at rest.

A velocity–time graph has time on the horizontal axis and velocity on the vertical axis. Its slope gives acceleration. A horizontal line means constant velocity and zero acceleration. If the line slopes upward, acceleration is positive; if it slopes downward, acceleration is negative. The signed area between the line and the time axis gives displacement. Area below the axis counts as negative displacement.

A graph can be read in sections. Find the slope between two points by dividing the change in the vertical quantity by the change in time. For a velocity–time graph, the area of a rectangle is base times height. A triangular area is one-half times base times height. Keep the sign when the graph is below the time axis.

v=\frac{\Delta x}{\Delta t},\quad a=\frac{\Delta v}{\Delta t},\quad \Delta x=\text{signed area under a velocity–time graph}

Position–time slope represents velocity.
Velocity–time slope represents acceleration.
Signed area under a velocity–time graph represents displacement.

3. Algebraic models for uniform and accelerated motion

For uniform motion, displacement equals constant velocity multiplied by elapsed time. Use this when velocity does not change. If the object begins at position

x_i

, its final position is the initial position plus its displacement.

For constant acceleration, the change in velocity is acceleration multiplied by time. The average velocity is the midpoint of initial and final velocity when acceleration is constant. Multiplying that average velocity by elapsed time gives displacement. These relationships connect the same quantities shown by the graphs.

Before substituting, list the known values and the unknown. Convert values to SI units if needed. Include units in substitutions, then round the final result to a sensible number of significant figures. A final check should confirm that the units match the requested quantity and that the sign agrees with the direction of motion.

\Delta x=vt,\quad v_f=v_i+at,\quad \Delta x=\frac{v_i+v_f}{2}t,\quad \Delta x=v_it+\frac{1}{2}at^2

Use the uniform-motion relationship only when velocity is constant.
Use constant-acceleration relationships only when acceleration remains constant.
Choose a relationship that includes the known values and the requested unknown.

Worked example

Uniform motion from position and time

A cart is the system. It moves east from position

x_i=2.0\ \mathrm{m}

to

x_f=14.0\ \mathrm{m}

in

4.0\ \mathrm{s}

. East is positive. Find its constant velocity.

Find displacement
Displacement is final position minus initial position. A positive result points east under the stated sign convention.
$\Delta x=(14.0\ \mathrm{m})-(2.0\ \mathrm{m})=+12.0\ \mathrm{m}$
Apply the uniform-motion model
The cart’s velocity is constant, so divide displacement by elapsed time. The units reduce to metres per second.
$v=\frac{\Delta x}{\Delta t}=\frac{+12.0\ \mathrm{m}}{4.0\ \mathrm{s}}=+3.0\ \mathrm{m/s}$

Answer: The cart’s velocity is

3.0\ \mathrm{m/s}

east.

Check: The units are metres per second, as required for velocity. The positive sign means east. A cart travelling at this speed for

4.0\ \mathrm{s}

covers

12.0\ \mathrm{m}

, matching the position change.

Worked example

Displacement from a velocity–time graph

A cyclist is the system. For

6.0\ \mathrm{s}

, the velocity–time graph is a straight line rising from

2.0\ \mathrm{m/s}

east to

8.0\ \mathrm{m/s}

east. Find the acceleration and displacement.

Use the graph’s slope
The slope of a velocity–time graph gives acceleration. The cyclist’s velocity increases in the positive, eastward direction.
$a=\frac{(8.0-2.0)\ \mathrm{m/s}}{6.0\ \mathrm{s}}=+1.0\ \mathrm{m/s^2}$
Find the signed area
The graph forms a trapezoid above the time axis. Its area equals average velocity times time, so it gives positive displacement.
$\Delta x=\frac{2.0\ \mathrm{m/s}+8.0\ \mathrm{m/s}}{2}(6.0\ \mathrm{s})=+30\ \mathrm{m}$

Answer: Acceleration is

1.0\ \mathrm{m/s^2}

east, and displacement is

30\ \mathrm{m}

east.

Check: The slope units are

(\mathrm{m/s})/\mathrm{s}=\mathrm{m/s^2}

. The area units are

(\mathrm{m/s})(\mathrm{s})=\mathrm{m}

. The displacement is reasonable: the cyclist travels for

6.0\ \mathrm{s}

at speeds between

2.0

and

8.0\ \mathrm{m/s}

, so an average speed of

5.0\ \mathrm{m/s}

gives

30\ \mathrm{m}

.

Worked example

Finding final velocity and displacement

A train is the system. It moves west at

12.0\ \mathrm{m/s}

and accelerates west at

0.80\ \mathrm{m/s^2}

for

5.0\ \mathrm{s}

. West is positive. Find its final velocity and displacement.

Set signed values
Because west is positive, both the initial velocity and acceleration are positive. The time interval is also known.
$v_i=+12.0\ \mathrm{m/s},\quad a=+0.80\ \mathrm{m/s^2},\quad t=5.0\ \mathrm{s}$
Calculate final velocity
For constant acceleration, velocity changes by acceleration multiplied by time. The positive result means west.
$v_f=v_i+at=12.0\ \mathrm{m/s}+(0.80\ \mathrm{m/s^2})(5.0\ \mathrm{s})=+16\ \mathrm{m/s}$
Calculate displacement
With constant acceleration, average velocity is the mean of initial and final velocity. Multiply it by time to find displacement.
$\Delta x=\frac{12.0\ \mathrm{m/s}+16.0\ \mathrm{m/s}}{2}(5.0\ \mathrm{s})=+70\ \mathrm{m}$

Answer: The train’s final velocity is

16\ \mathrm{m/s}

west, and its displacement is

70\ \mathrm{m}

west.

Check: Acceleration times time has units of velocity, and average velocity times time has units of displacement. Both answers are positive, matching west. The train speeds up from

12.0

to about

16\ \mathrm{m/s}

, so a displacement near

14\ \mathrm{m/s}\times5.0\ \mathrm{s}=70\ \mathrm{m}

is reasonable.

Common mistakes and how to avoid them

Treating a negative velocity as a negative speed.

Correction: Speed is a non-negative scalar. A negative velocity means motion in the direction opposite to the chosen positive direction.

Reading the height of a position–time graph as velocity.

Correction: The graph’s slope gives velocity. The height gives position.

Using the area under a position–time graph as displacement.

Correction: Displacement comes from signed area under a velocity–time graph. Position–time slope gives velocity.

Using a constant-acceleration relationship when acceleration changes.

Correction: These algebraic relationships require constant acceleration over the stated time interval.

Dropping units or changing positive direction partway through a solution.

Correction: Write units during substitution and use one stated sign convention for every vector value.

Lesson summary

State the system and positive direction before solving.
Uniform motion has constant velocity; constant acceleration means velocity changes at a steady rate.
Position–time slope gives velocity, and velocity–time slope gives acceleration.
Signed area under a velocity–time graph gives displacement.
Check units, signs, significant figures, and whether the result is physically reasonable.

Check your understanding

Question 1

A runner’s position changes from

3.0\ \mathrm{m}

to

-5.0\ \mathrm{m}

in

2.0\ \mathrm{s}

. What is the average velocity?

$+4.0\ \mathrm{m/s}$
$-4.0\ \mathrm{m/s}$
$-1.0\ \mathrm{m/s}$
$+8.0\ \mathrm{m/s}$

Show answer and explanation

-4.0\ \mathrm{m/s}

Displacement is

-5.0\ \mathrm{m}-3.0\ \mathrm{m}=-8.0\ \mathrm{m}

. Dividing by

2.0\ \mathrm{s}

gives

-4.0\ \mathrm{m/s}

.

Question 2

A velocity–time graph is a horizontal line at

-3.0\ \mathrm{m/s}

for

4.0\ \mathrm{s}

. What is the displacement?

$+12\ \mathrm{m}$
$-0.75\ \mathrm{m}$
$-12\ \mathrm{m}$
$0\ \mathrm{m}$

Show answer and explanation

-12\ \mathrm{m}

The signed rectangular area is

(-3.0\ \mathrm{m/s})(4.0\ \mathrm{s})=-12\ \mathrm{m}

. The negative sign gives the direction.

Question 3

An object has initial velocity

+5.0\ \mathrm{m/s}

and acceleration

-2.0\ \mathrm{m/s^2}

for

3.0\ \mathrm{s}

. What is its final velocity?

$+11\ \mathrm{m/s}$
$-1.0\ \mathrm{m/s}$
$+1.0\ \mathrm{m/s}$
$-6.0\ \mathrm{m/s}$

Show answer and explanation

+1.0\ \mathrm{m/s}

Using

v_f=v_i+at

, the result is

5.0\ \mathrm{m/s}+(-2.0\ \mathrm{m/s^2})(3.0\ \mathrm{s})=-1.0\ \mathrm{m/s}

? Recheck:

5.0-6.0=-1.0\ \mathrm{m/s}

, so the correct choice is

-1.0\ \mathrm{m/s}

.

Key terms

Displacement: Change in position, including direction.
Velocity: Displacement divided by elapsed time; it includes direction.
Acceleration: Change in velocity divided by elapsed time.
Uniform motion: Motion with constant velocity.
Constant acceleration: Acceleration that stays the same over a time interval.
Slope: Change in the vertical graph quantity divided by change in the horizontal graph quantity.
Signed area: Graph area counted as positive above the horizontal axis and negative below it.

Continue through SPH3U

View the complete SPH3U Ontario Grade 11 Physics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation B2.7. 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.

Official curriculum reference

Report a correction or ask a question

B2.7 · Solve uniform and accelerated motion problems graphically and algebraically

Ontario Grade 11 Physics

Reading motion graphs and choosing algebraic relationships

What you will learn

1. Prerequisite bridge: quantities, signs, and units

2. What motion graphs show

3. Algebraic models for uniform and accelerated motion

Uniform motion from position and time

Displacement from a velocity–time graph

Finding final velocity and displacement

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH3U

About this lesson and its review