Acceleration is metres per second squared, not metres per second. Always convert times in minutes or hours to seconds first.

A train travels 4800\ m in 120\ s. Calculate its average speed. [2 marks]

A car slows uniformly from 30\ m/s to 12\ m/s in 6.0\ s. Calculate its acceleration. [2 marks]

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How do we describe how fast something moves and how quickly its motion changes?

Distance, displacement, speed, velocity and acceleration, the difference between scalars and vectors, and the equations linking them.

A CCEA GCSE Physics answer on distance and displacement, speed and velocity, acceleration, the difference between scalar and vector quantities, and how to use and rearrange the speed and acceleration equations.

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
Examples in context
Try this

What this dot point is asking

CCEA wants you to tell the difference between scalar and vector quantities, to define distance, displacement, speed, velocity and acceleration, and to use and rearrange the equations that link them. Speed and acceleration calculations appear in every Unit 1 paper, so the equations and units must be automatic.

The answer

Scalars and vectors

Distance is how far an object has travelled, regardless of direction. Displacement is the straight-line distance from start to finish in a stated direction. If you walk 3 m east and then 3 m west, the distance is 6 m but the displacement is 0 m.

Speed and velocity

Speed is how fast distance is covered. Velocity is speed in a stated direction, so it is a vector.

The average speed uses the total distance over the total time; the instantaneous speed is the speed at one moment, for example the reading on a speedometer.

Acceleration

Acceleration is the rate of change of velocity. An object accelerates if it speeds up, slows down (a negative acceleration, or deceleration) or changes direction.

Worked example: a sprinter

Acceleration and distance of a sprinter

A sprinter starts from rest and reaches $9.0\ \text{m/s}$ in $3.0\ \text{s}$ , then runs at that steady speed for a further $5.0\ \text{s}$ . Find the acceleration during the first phase and the distance covered at steady speed.

step Find the acceleration in the first phase

The sprinter starts from rest, so $u = 0$ .

a = \frac{v - u}{t} = \frac{9.0 - 0}{3.0} = 3.0\ \text{m/s}^2.

step Find the distance at steady speed

At steady speed the velocity is constant at $9.0\ \text{m/s}$ , so use $s = vt$ .

s = vt = 9.0 \times 5.0 = 45\ \text{m}.

So the sprinter accelerates at $3.0\ \text{m/s}^2$ and then covers $45\ \text{m}$ at top speed.

Examples in context

Example 1. A lift. A lift starting from rest reaches $2.0\ \text{m/s}$ in $4.0\ \text{s}$ , so its acceleration is $a = (2.0 - 0)/4.0 = 0.5\ \text{m/s}^2$ . As it slows to stop, the acceleration becomes negative even though the lift is still moving upward.

Example 2. A motorway journey. A car covers $90\ \text{km}$ in $1.0$ hour. Converting, $90\ \text{km} = 90000\ \text{m}$ and $1.0\ \text{h} = 3600\ \text{s}$ , so the average speed is $90000 / 3600 = 25\ \text{m/s}$ , which is about $90\ \text{km/h}$ . Always convert before substituting.

Try this

Q1. State one scalar quantity and one vector quantity. [2 marks]

Cue. Scalar: speed (or distance, mass, energy). Vector: velocity (or displacement, force, acceleration).

Q2. A train travels $4800\ \text{m}$ in $120\ \text{s}$ . Calculate its average speed. [2 marks]

Cue. $v = s/t = 4800/120 = 40\ \text{m/s}$ .

Q3. A car slows uniformly from $30\ \text{m/s}$ to $12\ \text{m/s}$ in $6.0\ \text{s}$ . Calculate its acceleration. [2 marks]

Cue. $a = (12 - 30)/6.0 = -3.0\ \text{m/s}^2$ (a deceleration).

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 cyclist travels 1500 m along a straight road in 100 s, then stops for 20 s. Calculate her average speed for the whole journey and explain why her velocity at the start and at the end of the stop is different.

Show worked answer →

Average speed uses the total distance and total time:

$\text{speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{1500}{100 + 20} = \dfrac{1500}{120} = 12.5\ \text{m/s}.$

While moving her velocity is about 15 m/s in the direction of travel; while stopped her velocity is 0 m/s. Velocity includes direction and instantaneous value, so it changes even though the average speed for the whole trip is a single number.

Markers reward total distance over total time, the value 12.5 m/s, and a statement that velocity has direction or is the instantaneous value while speed here is averaged.

CCEA style3 marksA car accelerates uniformly from 8.0 m/s to 26 m/s in 6.0 s. Calculate its acceleration and state its unit.

Show worked answer →

Acceleration is the change in velocity divided by the time taken:

$a = \dfrac{v - u}{t} = \dfrac{26 - 8.0}{6.0} = \dfrac{18}{6.0} = 3.0\ \text{m/s}^2.$

Markers reward the change in velocity (18 m/s), dividing by the time, the value 3.0, and the unit m/s squared.

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Sources & how we know this

CCEA GCSE Physics specification — CCEA (2017)