Skip to main content
EnglandPhysicsSyllabus dot point

How do we describe and calculate motion using distance, speed and velocity?

Distance, time and velocity: distance and displacement, speed and velocity, the speed equation, and interpreting distance-time graphs.

A focused answer to AQA GCSE Physics 4.5.6, covering the difference between distance and displacement and between speed and velocity, the speed equation, typical everyday speeds, and how to read and use distance-time graphs.

Generated by Claude Opus 4.88 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Distance and displacement, speed and velocity
  3. The speed equation
  4. Distance-time graphs
  5. Try this

What this dot point is asking

AQA wants you to distinguish distance from displacement and speed from velocity, use the speed equation, recall typical everyday speeds, and interpret distance-time graphs including finding speed from the gradient.

Distance and displacement, speed and velocity

An object can have a constant speed but a changing velocity if it changes direction, for example moving in a circle. This is a favourite AQA point: a car going round a roundabout at a steady 30 mph30\,mph has a constant speed but a constantly changing velocity, and because its velocity is changing it is accelerating, even though the speedometer reading is fixed. The acceleration is caused by a resultant force directed towards the centre of the circle. The distinction between scalars (size only) and vectors (size and direction) underlies all of this: distance and speed are scalars, while displacement, velocity and acceleration are vectors.

The speed equation

These rough values matter because some exam questions ask you to estimate, for example, how long a journey takes or whether a calculated speed is sensible. The actual speed of a moving object depends on factors such as age, fitness, terrain and the wind, so AQA only expects approximate figures. For comparison, a typical car on a motorway travels at roughly 30 m/s30\,m/s and a commercial aircraft cruises at around 250 m/s250\,m/s. Being able to place a calculated answer against these benchmarks is a quick way to spot an arithmetic error in an exam.

The speed equation gives the average speed over a journey when you use the total distance and the total time. The instantaneous speed, the speed at one moment, can be different, for example a car that speeds up and slows down has a varying instantaneous speed but a single average speed for the trip. AQA expects you to keep this distinction clear and to convert units carefully, for example changing minutes to seconds or kilometres to metres before substituting.

Distance-time graphs

Reading a distance-time graph is a common exam task. A straight sloping line shows constant speed (the steeper the slope, the faster), a flat horizontal line shows the object is stationary (distance not changing), and a line sloping back down to zero shows the object returning to its start. A curve that gets steeper shows the object speeding up, and a curve that levels off shows it slowing down. To find the speed at a specific instant on a curve, draw a tangent to the curve at that point and calculate the gradient of the tangent.

Try this

Q1. State the difference between speed and velocity. [2 marks]

  • Cue. Speed is a scalar (size only); velocity is speed in a given direction (a vector).

Q2. A car travels 300 m300\,m in 15 s15\,s. Calculate its speed. [2 marks]

  • Cue. v=st=30015=20 m/sv = \dfrac{s}{t} = \dfrac{300}{15} = 20\,m/s.

Exam-style practice questions

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

AQA 20184 marksA cyclist travels 1500 m1500\,\text{m} along a straight road in 5.05.0 minutes. Calculate the cyclist's average speed in metres per second, and explain how you could find the speed of the cyclist at a single instant from a distance-time graph.
Show worked answer β†’

First convert the time to seconds: 5.05.0 minutes is 300 s300\,\text{s} (1 mark, a common slip). Then use v=s/t=1500/300=5.0 m/sv = s/t = 1500 / 300 = 5.0\,\text{m/s} (2 marks). To find the speed at a single instant from a distance-time graph, you draw a tangent to the curve at that point and calculate the gradient of the tangent (change in distance divided by change in time), because the gradient of a distance-time graph is the speed (1 mark). Markers reward the time conversion, the correct average speed, and the tangent-gradient method for an instantaneous speed.

AQA 20213 marksExplain the difference between distance and displacement, and explain why an object moving at a constant speed around a circular track has a continuously changing velocity.
Show worked answer β†’

Distance is the total length of the path travelled and is a scalar, so it has size only (1 mark). Displacement is the straight-line distance from the start to the finish together with its direction, so it is a vector (1 mark). An object moving around a circular track at constant speed has a continuously changing velocity because velocity is a vector that includes direction, and the direction of motion is constantly changing as the object goes round, even though the size of the speed stays the same (1 mark). Markers reward the scalar-vector distinction and the recognition that a changing direction means a changing velocity.

Related dot points

Sources & how we know this