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Motion and graphs

Describe motion using distance-time and speed-time graphs, speed, acceleration and constant-acceleration equations.

12 free questions · Notes · Worked solutions

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What you will practise

  • Calculate speed from distance and time.
  • Interpret gradients and areas on motion graphs.
  • Calculate acceleration from velocity change.
  • Apply v=u+at for constant acceleration.

Revision notes

Speed and velocity

Average speed=distance/time. Velocity includes direction, so a change in velocity can mean a change in speed or direction. Convert units consistently before calculating.

Distance-time graphs

The gradient of a distance-time graph gives speed. A horizontal segment means the object is stationary; a steeper gradient means greater speed.

Speed-time graphs

The gradient of a velocity-time graph gives acceleration. The area under a speed-time graph gives distance travelled when motion remains in one direction.

Constant acceleration

For constant acceleration, v=u+at, where u is initial velocity, v final velocity, a acceleration and t time. State direction and sign where needed; this relation assumes a remains constant.

A worked example

A cyclist accelerates uniformly from 4 m/s to 16 m/s in 6 s. Find the acceleration and the distance travelled in this time, assuming the speed-time graph is a straight line.

  1. a=(v−u)/t=(16−4)/6=2 m/s².
  2. Distance is the area under the speed-time graph.
  3. Average speed=(4+16)/2=10 m/s.
  4. Distance=10×6=60 m.

2 m/s²; 60 m

Try it yourself

These are original practice questions, not official past-paper questions. Practice points count correct final answers; working is not automatically assessed.

Choose a question to begin.
Question 1 · Standard1 practice point

A distance-time graph is horizontal. What is the object's motion?

Need a hint?

A horizontal distance-time graph has zero gradient.

Worked solution
  1. Zero gradient means distance is not changing.
  2. The object is stationary.
Question 2 · Standard1 practice point

A car travels 150 m in 10 s at constant speed. What is its speed?

Need a hint?

Speed=distance divided by time.

Worked solution
  1. 150÷10=15 m/s.
Question 3 · Standard1 practice point

Velocity increases from 5 m/s to 17 m/s in 4 s. Find the acceleration.

Need a hint?

Acceleration is change in velocity divided by time.

Worked solution
  1. Change=17−5=12 m/s.
  2. Acceleration=12÷4=3 m/s².
Question 4 · Standard1 practice point

A velocity-time graph is a horizontal line at 8 m/s for 5 s. What distance is travelled?

Need a hint?

Distance is the area under the graph.

Worked solution
  1. Area=8×5=40.
  2. Distance=40 m.
Question 5 · Foundation1 practice point

A runner covers 400 m in 80 s. Find the average speed in m/s.

Need a hint?

Divide distance by time.

Worked solution
  1. Speed=400÷80.
  2. Speed=5 m/s.
Question 6 · Standard2 practice points

A scooter's velocity changes from 3 m/s to 15 m/s in 4 s. Find its acceleration.

Need a hint?

Use a=(v−u)/t.

Worked solution
  1. Change in velocity=15−3=12 m/s.
  2. a=12÷4=3 m/s².
Question 7 · Standard2 practice points

A vehicle starts at 6 m/s and accelerates at 2 m/s² for 5 s. Find its final velocity.

Need a hint?

Use v=u+at.

Worked solution
  1. v=6+(2×5).
  2. v=16 m/s.
Question 8 · Standard2 practice points

A velocity-time graph rises uniformly from 0 to 12 m/s over 4 s. Find the distance travelled.

Need a hint?

The area is a triangle.

Worked solution
  1. Area=½×base×height.
  2. Distance=½×4×12=24 m.
Question 9 · Foundation1 practice point

A train travels at a constant 20 m/s for 30 s. Find the distance travelled.

Need a hint?

Distance=speed×time.

Worked solution
  1. d=20×30.
  2. d=600 m.
Question 10 · Standard2 practice points

A car slows uniformly from 25 m/s to 5 m/s in 10 s. Find its acceleration, including its sign.

Need a hint?

Use final minus initial velocity over time.

Worked solution
  1. a=(5−25)/10.
  2. a=−20/10=−2 m/s².
Question 11 · Standard2 practice points

A distance-time graph passes through (0 s, 0 m) and (8 s, 48 m). Find the constant speed represented.

Need a hint?

Its gradient is change in distance divided by change in time.

Worked solution
  1. Gradient=(48−0)/(8−0).
  2. Speed=6 m/s.
Question 12 · Stretch3 practice points

A velocity-time graph rises in a straight line from 4 m/s at 0 s to 16 m/s at 6 s. Find the distance travelled from 0 to 6 s.

Need a hint?

Use the trapezium area, or average velocity multiplied by time.

Worked solution
  1. Average velocity=(4+16)/2=10 m/s.
  2. Distance=10×6=60 m.

Coverage and sources

Selected IGCSE concepts, with a mix of introductory and more demanding questions. Check your own specification. AI drafted; independent teacher review is pending. Editorial details.

Cambridge IGCSE Physics 0625 syllabus

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