A runner covers 100\,m in 12.5\,s. Calculate the average speed. [2 marks]

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How are distance, speed and acceleration defined, and how do motion graphs describe movement?

Speed, velocity and acceleration, the equations of motion, and reading distance-time and velocity-time graphs.

A focused answer to WJEC GCSE Physics topic 2.1 on distance, speed and acceleration, covering the difference between speed and velocity, the acceleration equation, the equation of motion, and how to read distance-time and velocity-time graphs.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-14

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

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What this topic is asking
Speed, velocity and acceleration
The equation of motion
Reading motion graphs
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What this topic is asking

WJEC wants you to define speed, velocity and acceleration, use the equations of motion, and read distance-time and velocity-time graphs. This is topic 2.1 Distance, speed and acceleration in Unit 2 of WJEC GCSE Physics (3420).

Speed, velocity and acceleration

The difference between speed and velocity matters in Unit 2 because velocity carries a direction. An object moving in a circle at a steady speed has a constantly changing velocity, because its direction keeps changing, so it is accelerating even though its speed is constant. For most GCSE calculations, though, you work in a straight line and the numbers for speed and velocity are the same. Typical speeds are worth remembering: walking is roughly $1.5\,\text{m/s}$ , running about $3\,\text{m/s}$ , and a car in a town might travel at $13\,\text{m/s}$ (about $30\,\text{mph}$ ).

The equation of motion

When a question gives you three of the four quantities, choose $v = u + at$ and substitute. Watch the signs: if an object is slowing down, the acceleration is negative, so $v$ comes out smaller than $u$ . Higher candidates are expected to rearrange the equation to make any of the four quantities the subject; Foundation candidates are given it in the form they need.

Reading motion graphs

Graph questions are common in Unit 1 and Unit 2. The most reliable method is to label what each axis shows, then decide whether you need a gradient (a rate of change) or an area (a total). For a distance-time graph the gradient over a section is $\dfrac{\text{change in distance}}{\text{change in time}}$ , which is the speed for that section. For a velocity-time graph the gradient is the acceleration, and the area beneath gives the distance.

Distance from a velocity-time graph

A cyclist accelerates uniformly from rest to $12\,\text{m/s}$ over $6.0\,\text{s}$ , then travels at $12\,\text{m/s}$ for a further $10\,\text{s}$ . Find the total distance.

step 1: Distance during the acceleration

On a velocity-time graph this is a triangle of base $6.0\,\text{s}$ and height $12\,\text{m/s}$ . Area $= \tfrac{1}{2} \times 6.0 \times 12 = 36\,\text{m}$ .

step 2: Distance at constant velocity

This is a rectangle of base $10\,\text{s}$ and height $12\,\text{m/s}$ . Area $= 10 \times 12 = 120\,\text{m}$ .

step 3: Add the areas

Total distance $= 36 + 120 = 156\,\text{m}$ .

step 4: State the answer

The cyclist travels $156\,\text{m}$ . The distance is always the area under the velocity-time graph, found by adding the simple shapes.

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Q1. A runner covers $100\,\text{m}$ in $12.5\,\text{s}$ . Calculate the average speed. [2 marks]

Cue. $\text{speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{100}{12.5} = 8.0\,\text{m/s}$ .

Q2. State what the gradient of a distance-time graph represents. [1 mark]

Cue. The speed of the object.

Exam-style practice questions

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

WJEC 20193 marksA car accelerates uniformly from rest to

24\,\text{m/s}

8.0\,\text{s}

. Calculate its acceleration.

Show worked answer →

A topic 2.1 Calculate question. Use $a = \dfrac{\Delta v}{t}$ where the change in velocity is $24 - 0 = 24\,\text{m/s}$ (1 mark) and $t = 8.0\,\text{s}$ . Substitute: $a = \dfrac{24}{8.0}$ (1 mark) $= 3.0\,\text{m/s}^2$ (1 mark for the answer with units). Markers reward the change in velocity, the substitution and the unit metres per second squared. A common error is to forget that the car starts from rest, so the change in velocity is the full final velocity.

WJEC 20214 marksDescribe how you would find the acceleration and the distance travelled from a velocity-time graph.

Show worked answer →

A topic 2.1 Describe question. The acceleration is the gradient (slope) of the velocity-time graph, found by dividing the change in velocity on the vertical axis by the time taken on the horizontal axis (2 marks). The distance travelled is the area under the graph (2 marks); for a straight section this is the area of a rectangle or triangle. Markers reward gradient for acceleration and area for distance. A common error is to swap the two, reading the area as the acceleration.

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Sources & how we know this

WJEC GCSE Physics specification (3420) from 2016 — WJEC (2016)