Read gradients and areas on motion graphs
Relate distance–time, displacement–time, speed–time and velocity–time graphs to the quantities they represent.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 2. The falling-object activity is an online description and calculation practice; it does not provide supervised laboratory competence. Full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use SI units and signed values, and recall that velocity includes direction.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does the gradient of a displacement–time graph represent?
What you will learn
- Represent motion quantities graphically
- Find displacement from signed area under a velocity–time graph
- Find velocity from a displacement–time gradient
- Find acceleration from a velocity–time gradient
A graph's slope is a rate of change
A distance–time graph plots cumulative path length; its gradient gives speed. Distance does not decrease when an object reverses. A displacement–time graph is signed relative to an origin and its gradient gives velocity. For a curved displacement graph, a tangent gradient gives instantaneous velocity.
A speed–time graph shows speed magnitude and cannot show direction by itself. A velocity–time graph preserves direction; its gradient gives acceleration. A horizontal velocity segment is constant velocity, while a horizontal speed segment is constant speed. A negative velocity–time gradient means acceleration opposite to the chosen positive direction.
Area under velocity gives signed displacement
The signed area between a velocity–time graph and the time axis gives displacement. Areas above the axis are positive and areas below are negative. The area under a speed–time graph gives total distance because speed is non-negative; it does not show the direction of travel.
Use the shape under the graph: rectangle area is base × height and triangle area is half × base × height. Label axes and units so a gradient or area has the correct physical unit. A distance or displacement graph's gradient has units of distance per time; the area under a velocity–time graph has units of displacement.
Pause and explain
What does the signed area under a velocity–time graph give?
Worked example
Velocity rises uniformly from 4 m s⁻¹ to 16 m s⁻¹ in 6 s. Find acceleration and displacement during the interval.
Show the worked solution
- Velocity–time gradient = (16 − 4)/6 = 2.0 m s⁻².
- The area is a trapezium: average velocity × time = [(4 + 16)/2] × 6.
- Check that m s⁻¹ × s gives metres.
Answer Acceleration 2.0 m s⁻²; displacement 60 m.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A displacement–time graph passes through (t=2 s, s=5 m) and (t=8 s, s=23 m) on a straight segment. Find velocity.
Give me a hint
Use gradient = change in displacement ÷ change in time.
Compare my reasoning
- Change in displacement = 23 − 5 = 18 m.
- Change in time = 8 − 2 = 6 s.
- Gradient = 18/6.
3.0 m s⁻¹.
Look for these in your work
- Correct coordinate differences
- Gradient interpreted as velocity
- Unit m s⁻¹
Velocity increases uniformly from 2 m s⁻¹ to 10 m s⁻¹ over 4 s. Find acceleration and displacement.
Give me a hint
Use the gradient for acceleration and trapezium area for displacement.
Compare my reasoning
- Acceleration = (10 − 2)/4 = 2 m s⁻².
- Average velocity = (2 + 10)/2 = 6 m s⁻¹.
- Displacement = 6 × 4.
Acceleration 2 m s⁻²; displacement 24 m.
Look for these in your work
- Gradient and area meanings separated
- Trapezium area calculated
- Units correct
An object moves at +5 m s⁻¹ for 3 s, then at −2 m s⁻¹ for 4 s. Find its displacement and total distance.
Give me a hint
Use signed areas for displacement; add the magnitudes for distance.
Compare my reasoning
- Positive area = 5 × 3 = 15 m.
- Negative area = −2 × 4 = −8 m.
- Displacement = 15 − 8; distance = 15 + 8.
Displacement 7 m in the positive direction; total distance 23 m.
Look for these in your work
- Signed areas combined for displacement
- Area magnitudes added for distance
Common mistakes
- Using the gradient of a velocity–time graph as velocity instead of acceleration
- Treating the signed area as total distance after a reversal
- Reading a graph without checking its axis scale or units
What can you explain now?
What does the gradient of a displacement–time graph represent?
Compare with the explanation
Velocity
Gradient is change in displacement per change in time.
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Make it stick
Which graph operation gives velocity, acceleration and displacement, respectively?
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