A velocity-time graph shows constant 12\ m/s for 20\ s. Find the distance travelled. [2 marks]

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How do motion graphs show speed, acceleration and distance travelled?

Interpreting distance-time and velocity-time graphs, finding speed and acceleration from gradients, and distance from the area under a velocity-time graph.

A CCEA GCSE Physics answer on reading distance-time and velocity-time graphs, finding speed and acceleration from gradients, and calculating distance travelled from the area under a velocity-time graph.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
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What this dot point is asking

CCEA wants you to read distance-time and velocity-time graphs: to find speed from the gradient of a distance-time graph, acceleration from the gradient of a velocity-time graph, and distance from the area under a velocity-time graph. Graph questions are worth several marks in every Unit 1 paper.

The answer

Distance-time graphs

On a distance-time graph, time is on the horizontal axis and distance on the vertical axis.

To find the speed, pick two points on the line and use

\text{speed} = \frac{\text{change in distance}}{\text{change in time}}.

Velocity-time graphs

On a velocity-time graph, time is on the horizontal axis and velocity on the vertical axis. These graphs carry two pieces of information.

Finding acceleration and distance

The acceleration is found from the gradient:

a = \frac{\text{change in velocity}}{\text{time}} = \frac{v - u}{t}.

The distance is the area under the graph. Split the shape into rectangles and triangles, find each area, and add them.

Worked example: reading a velocity-time graph

Acceleration, deceleration and distance

A van accelerates uniformly from rest to $15\ \text{m/s}$ in $5.0\ \text{s}$ , holds $15\ \text{m/s}$ for $10\ \text{s}$ , then decelerates uniformly to rest in $3.0\ \text{s}$ . Find the deceleration in the last phase and the total distance.

step Find the deceleration in the last phase

a = \frac{v - u}{t} = \frac{0 - 15}{3.0} = -5.0\ \text{m/s}^2,

a deceleration of

5.0\ \text{m/s}^2

step Find the distance in each phase from the area

Acceleration phase (triangle): $\tfrac{1}{2} \times 5.0 \times 15 = 37.5\ \text{m}$ .
Constant phase (rectangle): $15 \times 10 = 150\ \text{m}$ .
Deceleration phase (triangle): $\tfrac{1}{2} \times 3.0 \times 15 = 22.5\ \text{m}$ .

step Add the areas for the total distance

s = 37.5 + 150 + 22.5 = 210\ \text{m}.

Examples in context

Example 1. A journey with a stop. A distance-time graph that rises, flattens, then rises again shows the object moving, stopping (the flat part), then moving again. The steeper of the two sloping sections is the faster part of the journey.

Example 2. A bus pulling away. A velocity-time graph that curves upward and gets steeper shows increasing acceleration; one that curves and levels off shows the bus approaching a steady cruising speed as the driving force balances resistance.

Try this

Q1. What does the gradient of a velocity-time graph represent? [1 mark]

Cue. The acceleration.

Q2. A velocity-time graph shows constant $12\ \text{m/s}$ for $20\ \text{s}$ . Find the distance travelled. [2 marks]

Cue. Area $= 12 \times 20 = 240\ \text{m}$ .

Q3. A car covers $400\ \text{m}$ at constant speed shown by a straight distance-time line spanning $25\ \text{s}$ . Find its speed. [2 marks]

Cue. Speed $=$ gradient $= 400/25 = 16\ \text{m/s}$ .

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA style4 marksA velocity-time graph shows a car accelerating uniformly from rest to 20 m/s in 8.0 s, then travelling at 20 m/s for 12 s. Calculate the acceleration in the first phase and the total distance travelled.

Show worked answer →

Acceleration is the gradient of the velocity-time graph in the first phase:

$a = \dfrac{v - u}{t} = \dfrac{20 - 0}{8.0} = 2.5\ \text{m/s}^2.$

Distance is the area under the graph. The first phase is a triangle, the second a rectangle:

$s_1 = \tfrac{1}{2} \times 8.0 \times 20 = 80\ \text{m}, \qquad s_2 = 20 \times 12 = 240\ \text{m}.$

Total distance $= 80 + 240 = 320\ \text{m}.$

Markers reward gradient for acceleration, the area method for distance, the two areas, and the total 320 m.

CCEA style3 marksDescribe how you would find the speed of an object from a straight section of a distance-time graph, and state what a horizontal line on such a graph represents.

Show worked answer →

The speed is the gradient of the line: choose two points on the straight section, then divide the change in distance by the change in time, $\text{speed} = \dfrac{\Delta \text{distance}}{\Delta \text{time}}$ .

A horizontal line means the distance is not changing, so the object is stationary (at rest).

Markers reward gradient equals change in distance over change in time, and that a horizontal line means stationary.

Related dot points

Sources & how we know this

CCEA GCSE Physics specification — CCEA (2017)