What does the gradient of a velocity–time graph give?

acceleration. Change in velocity divided by time.

What does the area under a velocity–time graph give?

displacement. Velocity multiplied by time.

What does the area under an acceleration–time graph give?

change in velocity. Acceleration multiplied by time.

A displacement–time graph is a curve getting steeper. What is the object doing?

speeding up. The gradient, which is the velocity, is increasing.

How do you find the velocity at one instant from a curved displacement–time graph?

draw a tangent and find its gradient. The gradient at a point is the gradient of the tangent there.

Graphs of motion · Cambridge A Level Physics 9702 revision

In plain words

A graph packs a whole journey into one picture, and two ideas unlock all of them: what the slope means, and what the area underneath means. If the line is curved, the slope at a point is the slope of the tangent there.

Three things to know

Displacement–time: the gradient is the velocity. Velocity–time: the gradient is the acceleration, and the area under the line is the displacement.
Acceleration–time: the area under the line is the change in velocity.
For a curved graph, draw a tangent at the point and find its gradient. Area below the time axis counts as negative.

Worked example

A velocity–time graph rises in a straight line from 0 to 12 m/s in 4.0 s, then stays at 12 m/s for 6.0 s. Find the displacement.

The area is a triangle and a rectangle.
Triangle: ½ × 4.0 × 12 = 24 m. Rectangle: 6.0 × 12 = 72 m.
Displacement = 96 m.

Tips and tricks

Use the largest triangle you can when measuring a gradient: it reduces the uncertainty.
Displacement can be negative. Distance can't. On a velocity–time graph, area below the axis is displacement backwards.

5 questions, about 2 minutes.

It lands in your notebook with its questions as flashcards.

Graphs of motion: 5 questions and answers

These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.

What does the gradient of a velocity–time graph give?
- displacement
- velocity
- acceleration (the answer)
- force
Change in velocity divided by time.
What does the area under a velocity–time graph give?
- displacement (the answer)
- acceleration
- speed
- time
Velocity multiplied by time.
What does the area under an acceleration–time graph give?
- displacement
- change in velocity (the answer)
- force
- acceleration
Acceleration multiplied by time.
A displacement–time graph is a curve getting steeper. What is the object doing?
- moving at constant velocity
- speeding up (the answer)
- slowing down
- stationary
The gradient, which is the velocity, is increasing.
How do you find the velocity at one instant from a curved displacement–time graph?
- divide the displacement by the time
- draw a tangent and find its gradient (the answer)
- find the area under the curve
- read the value off the axis
The gradient at a point is the gradient of the tangent there.

5 questions

Tap an answer and you’ll see straight away whether it’s right, and why.

Worksheet

3 questions, 7 marks. Write your answers on paper, then check them.

Graphs of motion

Cambridge International AS & A Level Physics 9702 · 7 marks · papermunch.org

Name ______________________________ Date ______________

A displacement–time graph is a straight line from 0 to 30 m over 6.0 s. Find the velocity.[2]
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5.0 m/s.
I got this one
A velocity–time graph falls in a straight line from 20 m/s to zero in 5.0 s. Find the acceleration and the displacement.[3]
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−4.0 m/s², and 50 m.
I got this one
Describe the velocity–time graph of a ball thrown vertically upwards and caught at the same height, taking upwards as positive.[2]
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A straight line sloping down with gradient −9.81 m/s², starting at a positive velocity, passing through zero at the top, and ending at an equal negative velocity.
I got this one

Answers: Graphs of motion

1. 5.0 m/s.
2. −4.0 m/s², and 50 m.
3. A straight line sloping down with gradient −9.81 m/s², starting at a positive velocity, passing through zero at the top, and ending at an equal negative velocity.