Learn9709Paper 4 · MechanicsKinematics of motion in a straight lineMotion graphs

Kinematics of motion in a straight line · Subtopic 3 of 4

Motion Graphs: Gradients and Areas

A velocity–time graph tells the whole story of a journey: its gradient is the acceleration and the area under it is the displacement. Most graph questions are really area questions in disguise.

Cambridge 9709 Paper 4Edexcel 9MA0 Paper 3AQA 7357 Paper 2≈ 30 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Read velocity from the gradient of a displacement–time graph.
Read acceleration from the gradient of a velocity–time graph, and displacement from its area.
Tell displacement from distance when the graph goes below the axis.
Set up and solve an equation when a time or speed on the graph is unknown.

Pause and think

A car drives at a steady $20\ \text{m s}^{-1}$ for $30$ s. Sketch its velocity–time graph. What shape is the region under it, and what does its area represent?

Section 2 of 14

Core idea

Start hereA bus journey in one picture

A bus speeds up, cruises, then brakes. A velocity–time graph shows the whole trip, and two features answer every question: the slope (how quickly speed changes) and the area (how far it went).

The big idea

Displacement–time: gradient = velocity. Velocity–time: gradient = acceleration; area = displacement.

1. Velocity–time graphs

Velocity–time graph of a bus journey

Distance

=\tfrac12(10)(12)+20(12)+\tfrac12(6)(12)=60+240+36=336

m. Or use the trapezium formula:

\tfrac12(20+36)(12)=336

Why area = displacement

$\text{constant }v\text{ for time }t:\ s=vt$
A rectangle of area $v\times t$ .
$\text{varying }v:\ \text{thin strips }v\,\delta t$
Each strip nearly constant speed.
$s=\sum v\,\delta t=\text{area}$
Add strips.

Worked exampleThe bus

0 to 12 m/s in 10 s; steady 20 s; stops in 6 s. Find accelerations, distance and average speed.

Goal: Values.
Strategy: Gradients and trapezium area.

$1.2,\ 0,\ -2\text{ m s}^{-2}$
$\frac{12}{10}$ , flat, $\frac{-12}6$ .
$\tfrac12(20+36)(12)=336\text{ m}$
Trapezium: parallel sides 20 and 36.
$\tfrac{336}{36}=9.33\text{ m/s}$
Total distance ÷ total time.

Answer

1.2

0

-2

;

336

9.33

m/s.

✓ Check

60+240+36=336

✓.

Worked exampleUnknown time from total distance

Accelerate 5 s to 20 m/s, steady $T$ s, stop in 10 s; total 500 m. Find $T$ .

Goal: $T$ .
Strategy: Area equation.

$\tfrac12(T+T+15)(20)=500$
Trapezium.
$2T+15=50\Rightarrow T=17.5$
Solve.

Answer

17.5

✓ Check

50+350+100=500

✓.

2. Displacement–time

The same journey as displacement–time

Steeper means faster. A straight section means constant velocity; a curve means the velocity is changing.

Edge caseBelow the axis Negative velocity means moving backwards; area below counts as negative displacement but positive distance.

Pause and think

Area under an acceleration–time graph?

Summary

$s$ – $t$ gradient: $v$ .
$v$ – $t$ gradient $a$ , area $s$ .

Gradient: Rate of change.

HistoryGraphs before graphs. In the 1330s, scholars at Merton College, Oxford worked out that a uniformly accelerating body covers the same distance as one moving at its average speed. Around 1350 the French bishop Nicole Oresme proved it with a diagram: speed drawn upwards against time, and distance as the area of the shape. That is a velocity–time graph, almost 300 years before Descartes' coordinates.

Fun factUsain Bolt's graph. In his

9.58

s world-record 100 m (Berlin, 2009), Bolt's velocity–time graph rises steeply for the first few seconds, levels off at a top speed of over

12\ \text{m s}^{-1}

between roughly 60 and 80 m, and dips slightly at the end. The area under it is exactly

100

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