Lesson focus: Velocity–time graphs encode acceleration in gradientHow steep a line on a graph is. and displacementHow far an object ends up from where it started, in a straight line and in a particular direction. in signed area beneath the graph.
Interpret velocity–time graphs, calculate acceleration from gradient and displacement from area.
Build on earlier learning
Motion graphs: interpreting journeysYear 8 interpreted distance–time journeys. This lesson changes both the vertical quantity and the meaning of gradient and area.
You will be successful when you can…
- Interpret graph sections.
- Calculate gradient acceleration.
- Calculate area displacement.
- Handle negative velocity correctly.
Words that carry the science
- velocity–time graph
- A graph showing velocity on the vertical axis against time.
- gradient
- Vertical change divided by horizontal change.
- area under the graph
- For velocity–time, the signed area representing displacement.
- constant acceleration
- Equal velocity change in equal time intervals.
- negative velocity
- Motion opposite to the chosen positive direction.
- displacement
- Signed change in position.
Gradient gives acceleration
01Gradient gives acceleration
On a velocity–time graphA graph that shows how the velocity of an object changes over time., gradient = change in velocity ÷ change in time, so it gives acceleration. A horizontal line means constant velocity and zero acceleration.
A straight sloping line means constant accelerationWhen velocity changes by the same amount every second.. A curve means acceleration changes, and the tangent gradientHow steep a line on a graph is. estimates instantaneous acceleration.
02Area gives displacement
The area between the velocity line and time axis has units m s⁻¹ × s = m, so it represents displacementHow far an object ends up from where it started, in a straight line and in a particular direction..
Area below the axis is negative displacement. Total distance requires adding the magnitudes of forward and backward areas rather than allowing cancellation.
03Axes determine meaning
On a distance–time graph, gradient is speed and area has no standard journey meaning. On a velocity–time graph, gradient is acceleration and area is displacement.
Always read labels and units before applying graph rules; memorising “gradient means speed” without checking axes causes errors.
Accelerating trolley
Set the acceleration and move the time slider. The curve gets steeper as the trolley speeds up.
Controls
Use gradient and area
Velocity rises uniformly from 0 to 12 m s⁻¹ in 4 s.
- Acceleration = (12 − 0) ÷ 4 = 3 m s⁻².
- The region is a triangle.
- Displacement = ½ × base × height.
- Displacement = ½ × 4 × 12 = 24 m.
How to make the reasoning visible: The same graph section gives acceleration from gradient and displacement from area.
Read a piecewise journey
Interpret successive straight sections of a velocity–time graph.
| Time / s | Velocity change / m s⁻¹ | Acceleration / m s⁻² | Section displacement / m |
|---|---|---|---|
| 0–4 | 0 to 12 | +3 | 24 |
| 4–9 | 12 to 12 | 0 | 60 |
| 9–12 | 12 to 0 | −4 | 18 |
| 12–14 | 0 to −4 | −2 | −4 |
Graph skills: SLAPUK and choosing a scaleVelocity–time graphs can go below zero.
What is total displacement?
What is total distance?
Check your understanding
Answer from memory, then check against the suggested answer.
What does velocity–time gradient represent?
What does signed area represent?
What does a line below the time axis mean?
A horizontal line is at −5 m s⁻¹. Describe the motion.
Why is total distance not always the signed area?
Exam Question Practice
Exam Question Practice
Mark points shown after marking
A car accelerates uniformly from 5 m/s to 17 m/s in 6 s, then travels at 17 m/s for 4 s.
Calculate the acceleration. Show your working.
(2)Calculate the total distance travelled in the 10 s. Show your working.
(4)Total for this question: 6 marks
Command word: For calculateA command word: work out a number answer, showing your working and the unit., state which graph feature you are using and show each section before adding.