B2.3 Derive and use constant-acceleration relationships in one dimension

Ontario Grade 11 Physics

Kinematics

Deriving and using motion relationships

A moving object can speed up, slow down, or change direction. In this lesson, we describe motion along one straight line when acceleration stays constant. We first choose a system, reference frame, and positive direction. Then we derive relationships from the definitions of acceleration and average velocity. These relationships let us find an unknown motion quantity when the other quantities are known.

What you will learn

Distinguish position, displacement, velocity, and acceleration, including their units and directions.
Explain how constant acceleration relates to velocity and time.
Use a velocity–time graph to connect displacement with average velocity.
Select and use a constant-acceleration relationship to solve one-dimensional motion problems.

1. Set up one-dimensional motion

The system is the object whose motion we are describing. A reference frame is the viewpoint used to measure its position and motion. For example, we can describe a cart’s motion relative to the floor. Choose one direction as positive before doing any calculations. Motion in the opposite direction is negative.

Position, written as

x

, tells where an object is relative to the chosen origin. Displacement, written as

\Delta x

, is the change in position: final position minus initial position. Both have the SI unit metre,

\mathrm{m}

. Displacement is a vector: it has a magnitude and a direction. A scalar, such as elapsed time, has magnitude but no direction.

Velocity describes how quickly position changes and includes direction. Average velocity is displacement divided by elapsed time. Acceleration describes how quickly velocity changes. In this lesson, the acceleration is constant, so velocity changes by equal amounts during equal time intervals. Velocity is measured in metres per second,

\mathrm{m/s}

, and acceleration in metres per second squared,

\mathrm{m/s^2}

.

Use the symbols

x_i

and

x_f

for initial and final position,

\Delta x

for displacement,

v_i

and

v_f

for initial and final velocity,

a

for acceleration, and

t

for elapsed time. The subscript

i

means initial;

f

means final. Choose a positive direction, write each known value with its sign and unit, and identify the unknown.

\Delta x=x_f-x_i

The sign of a vector gives its direction relative to the chosen positive direction.
Time is a scalar and is not assigned a direction.
For constant acceleration, velocity changes at a steady rate.

2. Derive the constant-acceleration relationships

Average acceleration is the change in velocity divided by elapsed time. When acceleration is constant, this definition can be rearranged to find final velocity. The equation works with positive or negative values: the signs record direction.

A velocity–time graph plots velocity vertically and time horizontally. With constant acceleration, the points form a straight line. The average velocity over the interval is the average of the initial and final velocities. Multiplying average velocity by elapsed time gives displacement. This is the area under the straight line on the velocity–time graph, but the average-velocity relationship is enough to calculate it.

Combining the velocity change relationship with the displacement relationship gives another useful equation. It connects velocity, acceleration, and displacement without requiring elapsed time. It is useful when time is not given and is not needed.

Each relationship applies only when acceleration is constant over the interval. Use consistent SI units. In particular, a negative velocity means motion in the negative direction; a negative acceleration means acceleration in that direction. A negative acceleration does not always mean an object is slowing down.

v_f=v_i+at,\qquad \Delta x=\frac{v_i+v_f}{2}t,\qquad v_f^2=v_i^2+2a\Delta x

Velocity–time graph slope represents acceleration in this constant-acceleration model.
The signed area under a velocity–time graph represents displacement.
Choose an equation that includes the known values and the unknown.

3. Choose an equation and check the result

Start by listing the known quantities and the unknown. For example, if time is not given, the relationship containing

v_f

,

v_i

,

a

, and

\Delta x

may be the best choice. If displacement is needed and both endpoint velocities and time are known, use average velocity.

Keep signs and units in the substitution. Do not replace a negative velocity or acceleration with its magnitude unless the question specifically asks for a magnitude. After calculating, state the direction in words or with a signed value and the chosen convention.

Check the units and whether the result makes sense. For example, multiplying velocity by time gives metres, the unit of displacement. A result with the wrong unit signals an equation or substitution error. Check also whether the object’s direction and speed change are consistent with the signs.

\mathrm{m/s^2}\times\mathrm{s}=\mathrm{m/s}

Write the governing relationship before substituting values.
Carry units through the calculation and round to a sensible number of significant figures.
Use the signs, units, and motion description to check reasonableness.

4. Common sign and model errors

A direction choice is not the same as a claim about which direction is naturally positive. Either direction can be positive, but all vector quantities must follow the same choice. If you reverse the positive direction, the signs of displacement, velocity, and acceleration reverse too.

Constant acceleration does not mean constant velocity. It means velocity changes steadily. If acceleration and velocity have the same sign, the object’s speed increases. If their signs differ, its speed decreases while it continues moving in its current direction. The equations describe the whole interval, so do not treat a changing velocity as though it were constant.

Do not confuse displacement with distance: displacement includes direction.
Do not assume negative acceleration always means slowing down.
Do not use a constant-acceleration relationship for an interval where acceleration is not constant.

Worked example

1. Find final velocity and displacement

A cart moves along a straight track. Take the direction of its motion as positive. It starts at

3.0\ \mathrm{m/s}

and accelerates at

2.0\ \mathrm{m/s^2}

for

4.0\ \mathrm{s}

. Find its final velocity and displacement.

Define the motion
The system is the cart, measured relative to the track. The positive direction is the cart’s initial direction. The known values are $v_i=+3.0\ \mathrm{m/s}$ , $a=+2.0\ \mathrm{m/s^2}$ , and $t=4.0\ \mathrm{s}$ . Find $v_f$ and $\Delta x$ .
Find final velocity
Use the constant-acceleration relationship for final velocity. Substitute the signed values and units.
$v_f=v_i+at=(3.0\ \mathrm{m/s})+(2.0\ \mathrm{m/s^2})(4.0\ \mathrm{s})=11\ \mathrm{m/s}$
Find displacement
The velocity increases steadily, so average velocity is the mean of the endpoint velocities. Multiply it by the elapsed time to get displacement.
$\Delta x=\frac{v_i+v_f}{2}t=\frac{3.0\ \mathrm{m/s}+11\ \mathrm{m/s}}{2}(4.0\ \mathrm{s})=28\ \mathrm{m}$

Answer: The final velocity is

+11\ \mathrm{m/s}

, in the positive direction. The displacement is

+28\ \mathrm{m}

.

Check: The units in the displacement calculation reduce to metres. Both velocity and acceleration are positive, so the cart speeds up in the positive direction. A displacement greater than the starting speed multiplied by time is reasonable because the cart gains speed.

Worked example

2. Find the stopping distance

A bicycle travels at

18\ \mathrm{m/s}

along a straight path. Take its direction of travel as positive. It slows with a constant acceleration of

-3.0\ \mathrm{m/s^2}

until it stops. Find the stopping time and displacement.

Define the motion
The system is the bicycle, and the positive direction is its initial direction of travel. The known values are $v_i=+18\ \mathrm{m/s}$ , $v_f=0\ \mathrm{m/s}$ , and $a=-3.0\ \mathrm{m/s^2}$ . First find time, then displacement.
Find stopping time
Rearrange the final-velocity relationship to isolate time. The negative acceleration reduces the positive velocity to zero.
$t=\frac{v_f-v_i}{a}=\frac{0\ \mathrm{m/s}-18\ \mathrm{m/s}}{-3.0\ \mathrm{m/s^2}}=6.0\ \mathrm{s}$
Find displacement
Use the average of the initial and final velocities over the stopping interval. The displacement remains positive because the bicycle moves forward until it stops.
$\Delta x=\frac{v_i+v_f}{2}t=\frac{18\ \mathrm{m/s}+0\ \mathrm{m/s}}{2}(6.0\ \mathrm{s})=54\ \mathrm{m}$

Answer: The bicycle takes

6.0\ \mathrm{s}

to stop and has a displacement of

+54\ \mathrm{m}

.

Check: The time is positive, and the displacement unit is metres. A negative acceleration with positive velocity reduces the bicycle’s speed. The average velocity is

9.0\ \mathrm{m/s}

, so covering

54\ \mathrm{m}

in

6.0\ \mathrm{s}

is consistent.

Worked example

3. Motion that reverses direction

A small object moves along a line. The positive direction is to the right. Its initial velocity is

-2.0\ \mathrm{m/s}

and its constant acceleration is

+1.5\ \mathrm{m/s^2}

for

4.0\ \mathrm{s}

. Find its final velocity and displacement.

Interpret the signs
The system is the object, and right is positive. Its initial velocity is negative, so it initially moves left. The positive acceleration points right. The known values are $v_i=-2.0\ \mathrm{m/s}$ , $a=+1.5\ \mathrm{m/s^2}$ , and $t=4.0\ \mathrm{s}$ .
Find final velocity
Use the velocity relationship with the signed initial velocity. The positive change in velocity is large enough for the object to finish moving right.
$v_f=v_i+at=(-2.0\ \mathrm{m/s})+(1.5\ \mathrm{m/s^2})(4.0\ \mathrm{s})=+4.0\ \mathrm{m/s}$
Find displacement
Use the average endpoint velocity over the interval. The signed result gives the net change in position, not the total distance travelled.
$\Delta x=\frac{v_i+v_f}{2}t=\frac{-2.0\ \mathrm{m/s}+4.0\ \mathrm{m/s}}{2}(4.0\ \mathrm{s})=+4.0\ \mathrm{m}$

Answer: The final velocity is

+4.0\ \mathrm{m/s}

, to the right. The displacement is

+4.0\ \mathrm{m}

, also to the right.

Check: The velocity changes from leftward to rightward, which is consistent with rightward acceleration. Average velocity is

+1.0\ \mathrm{m/s}

, giving

+4.0\ \mathrm{m}

in

4.0\ \mathrm{s}

. The displacement is plausible even though the object initially moved left.

Common mistakes and how to avoid them

Treating all speeds, velocities, and accelerations as positive magnitudes.

Correction: Velocity and acceleration are vectors. Include a sign based on the stated positive direction.

Assuming a negative acceleration always means an object is slowing down.

Correction: Compare the signs of velocity and acceleration. Opposite signs mean the speed decreases while the object moves in its current direction.

Using the initial velocity as though it stayed constant when calculating displacement.

Correction: For constant acceleration, use average velocity, or another suitable constant-acceleration relationship.

Reporting a displacement without a direction or sign.

Correction: Give the signed displacement or state its direction relative to the chosen positive direction.

Lesson summary

Define the system, reference frame, and positive direction before describing motion.
For constant acceleration, final velocity equals initial velocity plus acceleration multiplied by time.
Displacement equals average velocity multiplied by elapsed time; with constant acceleration, average velocity is the mean of the endpoint velocities.
The relationship $v_f^2=v_i^2+2a\Delta x$ is useful when time is not needed.
Include signs and SI units, then check direction, units, and reasonableness.

Check your understanding

Question 1

A cart has initial velocity

+5.0\ \mathrm{m/s}

and constant acceleration

-1.0\ \mathrm{m/s^2}

. What is its velocity after

3.0\ \mathrm{s}

?

$+8.0\ \mathrm{m/s}$
$+2.0\ \mathrm{m/s}$
$-2.0\ \mathrm{m/s}$
$+1.7\ \mathrm{m/s}$

Show answer and explanation

+2.0\ \mathrm{m/s}

Use

v_f=v_i+at

. Substitution gives

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

. The positive result means the cart is still moving in the positive direction.

Question 2

A cart has initial velocity

+5.0\ \mathrm{m/s}

and constant acceleration

-1.0\ \mathrm{m/s^2}

. What is its velocity after

3.0\ \mathrm{s}

?

$+8.0\ \mathrm{m/s}$
$+2.0\ \mathrm{m/s}$
$-2.0\ \mathrm{m/s}$
$+1.7\ \mathrm{m/s}$

Show answer and explanation

+2.0\ \mathrm{m/s}

Using

v_f=v_i+at

gives

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

. The positive sign means motion remains in the positive direction.

Key terms

System: The object whose motion is being described.
Reference frame: The viewpoint or location from which position and motion are measured.
Displacement: The change in position, including direction.
Velocity: The rate of change of position, including direction.
Acceleration: The rate of change of velocity.
Constant acceleration: Acceleration that has the same value and direction throughout the time interval.
Scalar: A quantity with magnitude but no direction, such as elapsed time.
Vector: A quantity with magnitude and direction, such as displacement, velocity, or acceleration.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation B2.3. 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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B2.3 · Derive and use constant-acceleration relationships in one dimension

Ontario Grade 11 Physics

Deriving and using motion relationships

What you will learn

1. Set up one-dimensional motion

2. Derive the constant-acceleration relationships

3. Choose an equation and check the result

4. Common sign and model errors

1. Find final velocity and displacement

2. Find the stopping distance

3. Motion that reverses direction

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through SPH3U

About this lesson and its review