AQA · GCSE · Combined Science
Course version: 8464 · Version from Sept 2018
The speed of an object can be calculated from the gradient of its…
Full specification wording
The distance–time relationship: The speed of an object can be calculated from the gradient of its distance–time graph.
A distance–time graph plots distance travelled on the vertical axis against time on the horizontal axis.
Checked against the AQA 8464 specification (opens in a new tab) on .
- Point 2 of 5
Study notes
From the checked teaching notes for this specification point.
The idea
A distance–time graph plots distance travelled on the vertical axis against time on the horizontal axis. For an object moving at constant speed, the line is straight and its gradient equals the speed. Gradient means vertical change divided by horizontal change, so choose two points on the line that are far apart, read the distance values and the time values, subtract to find each change, then divide distance change by time change. For example, if distance rises from 20 m to 80 m between 4 s and 16 s, the gradient is (80 − 20) ÷ (16 − 4) = 60 ÷ 12 = 5 m/s. A steeper line means a greater speed; a horizontal line means the object is stationary. Units come from dividing metres by seconds, giving m/s.
What earns the marks
- Identify the vertical axis as distance and the horizontal axis as time, and state that gradient means change in distance divided by change in time.
- Select two clearly separated points on the straight line, read their coordinates accurately from the axes, and subtract to find the changes rather than reading single values.
- Divide the change in distance by the change in time, including correct units such as m/s, and interpret a larger gradient as a greater speed.
- Explain that a horizontal section represents zero speed because the distance does not change as time increases, and that a straight line shows constant speed.
Common mistakes and how to put them right
Dividing the time change by the distance change.
Put it right by remembering gradient is rise over run, so distance change goes on top.
Reading a single coordinate and dividing distance by time without subtracting.
Put it right by using two points and finding the differences first.
Forgetting units or writing km/h when the axes use metres and seconds.
Put it right by deriving units from the axes, giving m/s here.
Exam tips
- Draw the two chosen points on the graph and mark the vertical and horizontal changes before calculating, so your method is visible.
- Use points on grid intersections where possible to reduce reading error, and show the subtraction step clearly.
- Check the size of your answer against the graph: a steeper line should give a larger speed than a shallower one.
Try the lesson
The start of the The distance–time relationship: The speed of an object can be calculated from the gradient of its distance–time graph. lesson, made from checked teaching notes for this specification point.
Have a go first
On a distance–time graph, a straight line rises by 20 m in a time of 4 s. What is the speed of the object?
Listen
A distance–time graph plots distance travelled on the vertical axis against time on the horizontal axis. For an object moving at constant speed, the line is straight and its gradient equals the speed. Gradient means vertical change divided by horizontal change, so choose two points on the line that are far apart, read the distance values and the time values, subtract to find each change, then divide distance change by time change. For example, if distance rises from 20 m to 80 m between 4 s and 16 s, the gradient is (80 − 20) ÷ (16 − 4) = 60 ÷ 12 = 5 m/s. A steeper line means a greater speed; a horizontal line means the object is stationary. Units come from dividing metres by seconds, giving m/s. A strong answer earns marks for these points. Identify the vertical axis as distance and the horizontal axis as time, and state that gradient means change in distance divided by change in time. Select two clearly separated points on the straight line, read their coordinates accurately from the axes, and subtract to find the changes rather than reading single values. Divide the change in distance by the change in time, including correct units such as m/s, and interpret a larger gradient as a greater speed. Explain that a horizontal section represents zero speed because the distance does not change as time increases, and that a straight line shows constant speed. A common slip: Dividing the time change by the distance change. Instead, remembering gradient is rise over run, so distance change goes on top. Exam tip: Draw the two chosen points on the graph and mark the vertical and horizontal changes before calculating, so your method is visible.
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