B2.2 Interpret position-time, velocity-time, and acceleration-time graphs

Ontario Grade 11 Physics

Kinematics

Interpreting position-time, velocity-time, and acceleration-time graphs

A motion graph shows how an object's motion changes over time. First define the system: the object being described. Define the reference frame: what the object's position is measured relative to. Then choose a positive direction. In this lesson, the system is a moving object on a straight path, measured relative to a point on that path. Right is positive; left is negative. Position, velocity, and acceleration are vectors, so their signs show direction relative to the chosen positive direction. Time is a scalar: it has size but no direction. Read the axes before interpreting a graph's shape.

What you will learn

Identify what each axis represents and use its units to interpret a motion graph.
Interpret slope on position-time and velocity-time graphs.
Interpret signed area on velocity-time and acceleration-time graphs.
Use acceleration-time graphs to describe how velocity changes.

1. Prerequisite bridge: axes, signs, and units

A graph has a horizontal axis and a vertical axis. Read the labels and units on both axes before interpreting the graph. Time is usually on the horizontal axis and increases from left to right.

Position,

x

, tells where an object is relative to the chosen reference point. Its SI unit is the metre,

\mathrm{m}

. Displacement,

\Delta x

, is the change in position. Velocity,

v

, describes the change in position over time and includes direction. Its SI unit is metres per second,

\mathrm{m/s}

. Acceleration,

a

, describes the change in velocity over time. Its SI unit is metres per second squared,

\mathrm{m/s^2}

.

A scalar has size but no direction. A vector has both size and direction. Time is a scalar. Position, displacement, velocity, and acceleration are vectors in one-dimensional motion. A positive or negative sign records direction relative to the chosen positive direction.

A straight segment's slope is its vertical change divided by its horizontal change. In motion graphs, the axis labels tell you what physical quantity that slope represents.

\frac{\Delta y}{\Delta x}

Choose and state a positive direction before interpreting signs.
Check both axis labels and units.
A negative vector value means a direction opposite to the chosen positive direction.

2. Position-time graphs

A position-time graph has time,

t

, on the horizontal axis and position,

x

, on the vertical axis. Each point gives the object's position at one time. A point at

t=3\,\mathrm{s}

and

x=6\,\mathrm{m}

means the object is six metres in the positive direction from the reference point after three seconds.

The slope of a position-time graph represents velocity. For a straight segment, divide the change in position by the time interval. A positive slope means motion in the positive direction; a negative slope means motion in the negative direction. A horizontal segment has zero slope, so the object is at rest during that interval. A steeper segment means a greater velocity magnitude.

A changing slope means the velocity is changing. Compare the steepness and sign of successive segments to describe how the motion changes. Do not mistake the graph's height for speed: its height gives position, while its slope gives velocity.

v=\frac{\Delta x}{\Delta t}

Vertical coordinate: position in metres.
Slope: velocity in metres per second.
Horizontal segment: position stays constant.

3. Velocity-time graphs

A velocity-time graph has time,

t

, on the horizontal axis and velocity,

v

, on the vertical axis. A point above the time axis represents positive velocity. A point below it represents negative velocity. A point on the axis represents zero velocity at that instant.

The slope of a velocity-time graph represents acceleration. A line rising from left to right has positive slope; a line falling from left to right has negative slope. A horizontal line means velocity is constant and acceleration is zero during that interval.

Negative acceleration does not always mean an object is slowing down. Compare the signs of velocity and acceleration. If they have opposite signs, the magnitude of velocity decreases. If they have the same sign, it increases.

The signed area between a velocity-time graph and the time axis represents displacement. Area above the axis is positive; area below it is negative. For a rectangular section, multiply velocity by elapsed time. For a triangular section, multiply one-half by the base and height. Add the signed areas to find total displacement. The area has units of metres.

a=\frac{\Delta v}{\Delta t}

Vertical coordinate: velocity in metres per second.
Slope: acceleration in metres per second squared.
Signed area: displacement in metres.

4. Acceleration-time graphs and a reading routine

An acceleration-time graph has time,

t

, on the horizontal axis and acceleration,

a

, on the vertical axis. A point above the axis means acceleration in the positive direction. A point below it means acceleration in the negative direction. A horizontal line can represent constant acceleration; a line on the time axis represents zero acceleration.

The signed area under an acceleration-time graph represents the change in velocity,

\Delta v

. Area above the axis gives a positive change in velocity, and area below gives a negative change. The area has units of metres per second. To find final velocity, combine the change in velocity with initial velocity, paying attention to their signs.

Use this routine for any motion graph: name the axes and units; state the positive direction; read the graph's value at a chosen time; then use slope or signed area only when that graph type gives it meaning. State whether your result is position, velocity, acceleration, displacement, or change in velocity.

Change in velocity equals the signed area under an acceleration-time graph.

Vertical coordinate: acceleration in metres per second squared.
Signed area: change in velocity in metres per second.
A graph's slope or area has meaning only when interpreted using its axis quantities.

Worked example

Finding velocity from a position-time graph

A cart moves along a straight track. Right is positive, and position is measured from a marked point. On a straight section of its position-time graph, the cart is at

x=2.0\,\mathrm{m}

at

t=1.0\,\mathrm{s}

and at

x=8.0\,\mathrm{m}

at

t=4.0\,\mathrm{s}

. Find its velocity on this section.

Set the system and direction
The system is the cart, and the reference point is the marked point on the track. Right is positive. The graph gives two positions and times; the unknown is the cart's velocity on the straight section.
Choose the graph relationship
The slope of a position-time graph gives velocity. Use the later value minus the earlier value for both position and time, so the signs stay consistent.
$v=\frac{\Delta x}{\Delta t}=\frac{x_2-x_1}{t_2-t_1}$
Substitute with units
The position change is positive, so the cart moves in the chosen positive direction.
$v=\frac{8.0\,\mathrm{m}-2.0\,\mathrm{m}}{4.0\,\mathrm{s}-1.0\,\mathrm{s}}=2.0\,\mathrm{m/s}$

Answer: The cart's velocity is

2.0\,\mathrm{m/s}

to the right.

Check: The units reduce to metres per second, as required for velocity. The positive sign matches rightward motion. A change of

6.0\,\mathrm{m}

over

3.0\,\mathrm{s}

is consistent with

2.0\,\mathrm{m/s}

.

Worked example

Reading slope and displacement from a velocity-time graph

A runner moves along a straight path. Forward is positive. From

t=0

to

t=4.0\,\mathrm{s}

, the velocity-time graph is a straight line from

+2.0\,\mathrm{m/s}

to

+6.0\,\mathrm{m/s}

. Find the acceleration and displacement during this interval.

Identify the system and graph information
The system is the runner, measured relative to the path. Forward is positive. The graph provides initial and final velocities and the elapsed time; acceleration and displacement are unknown.
Find acceleration from slope
The slope of a velocity-time graph gives acceleration. Both velocities are positive, and velocity increases, so acceleration should be positive.
$a=\frac{\Delta v}{\Delta t}=\frac{6.0\,\mathrm{m/s}-2.0\,\mathrm{m/s}}{4.0\,\mathrm{s}}=1.0\,\mathrm{m/s^2}$
Find displacement from signed area
The graph section is a trapezoid above the time axis. Its area is the average of the two velocity values multiplied by the time interval. Because the area is above the axis, displacement is positive.
$\Delta x=\frac{(2.0\,\mathrm{m/s}+6.0\,\mathrm{m/s})}{2}(4.0\,\mathrm{s})=16\,\mathrm{m}$

Answer: The acceleration is

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

forward, and the displacement is

16\,\mathrm{m}

forward.

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}

. Both answers are positive, matching the graph above the time axis. The displacement is reasonable: the runner travels for four seconds at speeds between

2.0

and

6.0\,\mathrm{m/s}

.

Worked example

Using an acceleration-time graph

A small cart moves along a straight floor. Right is positive. Its initial velocity is

+1.0\,\mathrm{m/s}

. For

3.0\,\mathrm{s}

, its acceleration-time graph is a horizontal line at

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

. Find the change in velocity and final velocity.

Set the system and known values
The system is the cart, and right is positive. The initial velocity, acceleration, and time interval are known. The graph's signed area gives the change in velocity.
Calculate the change in velocity
The area is a rectangle below the time axis, so it is negative. Its units are acceleration multiplied by time.
$\Delta v=(-2.0\,\mathrm{m/s^2})(3.0\,\mathrm{s})=-6.0\,\mathrm{m/s}$
Combine with initial velocity
Add the signed change to the initial velocity. A negative final value means the cart is moving left, opposite to the positive direction.
$v_f=v_i+\Delta v=+1.0\,\mathrm{m/s}-6.0\,\mathrm{m/s}=-5.0\,\mathrm{m/s}$

Answer: The change in velocity is

-6.0\,\mathrm{m/s}

. The final velocity is

5.0\,\mathrm{m/s}

to the left.

Check: The area units reduce to metres per second. The final velocity is negative, as expected after a substantial negative change from a small positive initial velocity. Its magnitude is greater than the initial speed, which is reasonable because the acceleration acts for the full interval.

Common mistakes and how to avoid them

Reading the height of a position-time graph as speed.

Correction: The height gives position. The slope gives velocity.

Calling every negative acceleration a decrease in speed.

Correction: Compare acceleration with velocity. Opposite signs mean speed decreases; the same sign means speed increases.

Treating all area under a velocity-time graph as positive.

Correction: Area below the time axis gives negative displacement. Keep the sign when adding graph sections.

Confusing acceleration-time area with displacement.

Correction: Area under an acceleration-time graph gives change in velocity. Its units are metres per second.

Ignoring the chosen positive direction.

Correction: State the direction convention first, then interpret every positive or negative vector value consistently.

Lesson summary

Read axis labels and units before interpreting a motion graph.
Position-time slope gives velocity: $v=\Delta x/\Delta t$ .
Velocity-time slope gives acceleration: $a=\Delta v/\Delta t$ .
Signed area under a velocity-time graph gives displacement.
Signed area under an acceleration-time graph gives change in velocity.
Use signs to describe direction relative to the chosen positive direction.

Check your understanding

Question 1

An object's position-time graph is horizontal at

x=-3.0\,\mathrm{m}

for two seconds. What does this show?

The object is at rest at a position three metres in the negative direction.
The object moves in the negative direction at $3.0\,\mathrm{m/s}$ .
The object has an acceleration of $-3.0\,\mathrm{m/s^2}$ .
The object moves in the positive direction.

Show answer and explanation

The object is at rest at a position three metres in the negative direction.

A horizontal position-time graph has zero slope, so velocity is zero. Its vertical coordinate gives the position, which is

-3.0\,\mathrm{m}

.

Question 2

A velocity-time graph is a horizontal line at

-4.0\,\mathrm{m/s}

. Which statement is correct?

The object has constant velocity in the negative direction and zero acceleration.
The object is at rest because its velocity is constant.
The object has positive acceleration.
The object moves in the positive direction at $4.0\,\mathrm{m/s}$ .

Show answer and explanation

The object has constant velocity in the negative direction and zero acceleration.

The vertical value gives velocity. Its negative sign indicates the negative direction, and the horizontal line has zero slope, so acceleration is zero.

Question 3

An acceleration-time graph shows a constant

+3.0\,\mathrm{m/s^2}

for

2.0\,\mathrm{s}

. What is the change in velocity?

$+6.0\,\mathrm{m/s}$
$+1.5\,\mathrm{m/s}$
$+5.0\,\mathrm{m/s}$
$-6.0\,\mathrm{m/s}$

Show answer and explanation

+6.0\,\mathrm{m/s}

The signed rectangular area is acceleration multiplied by time:

(+3.0\,\mathrm{m/s^2})(2.0\,\mathrm{s})=+6.0\,\mathrm{m/s}

. The units and positive sign are consistent.

Key terms

Reference frame: The chosen viewpoint or reference used to measure an object's position.
Positive direction: The direction assigned a positive sign for describing position changes, velocity, and acceleration.
Scalar: A quantity with size but no direction.
Vector: A quantity with both size and direction.
Slope: The change in a graph's vertical quantity divided by the change in its horizontal quantity.
Displacement: The change in position, including direction.
Signed area: Area counted as positive above an axis and negative below it.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation B2.2. It is a study resource, not an official curriculum publication.

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B2.2 · Interpret position-time, velocity-time, and acceleration-time graphs

Ontario Grade 11 Physics

Interpreting position-time, velocity-time, and acceleration-time graphs

What you will learn

1. Prerequisite bridge: axes, signs, and units

2. Position-time graphs

3. Velocity-time graphs

4. Acceleration-time graphs and a reading routine

Finding velocity from a position-time graph

Reading slope and displacement from a velocity-time graph

Using an acceleration-time graph

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