Plot distance-time graphs and calculate straight gradients
Read both axes and use a large gradient triangle to find a speed.
Exam scope and assessment notes
Core and Extended · Physics 0972, 2026–2028, section 1.2. Core includes speed, velocity direction, distance–time and speed–time interpretation, straight distance–time gradients, speed–time areas and near-Earth free-fall acceleration. Numerical acceleration and the resistance/terminal-velocity explanation are Supplement requirements, taught in separately labelled Extended lessons. Written practice is self-checked; cumulative assessment, supervised practical experience and human subject-teacher review remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Plot ordered coordinate pairs.
- Subtract two times and two distances.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which quantity belongs on the horizontal axis of a distance–time graph?
What you will learn
- Plot a distance–time record.
- Interpret rest and constant speed using gradient.
Plot the quantities rather than square counts
Put time on the horizontal axis and cumulative distance travelled on the vertical axis, labelled with units. Data (0 s, 0 m), (4 s, 12 m), (6 s, 12 m), (10 s, 20 m) give three straight sections if the motion is specified as constant within each interval. Read the values represented by each grid interval on both axes.
The first line rises at a constant rate, the middle is horizontal and the last rises less steeply. A horizontal cumulative-distance line means no further distance is travelled, so the object is at rest. A straight rising line means constant speed, not increasing speed merely because distance keeps growing.
A gradient is a rate of distance change
For a straight section, speed equals change in distance divided by change in time. From four to six seconds the numerator is zero. From six to ten seconds it is (20 − 12)/(10 − 6) = 2 m/s. Use endpoints on the same straight section; spanning a stop gives an interval average rather than that section's speed.
Choose a large triangle on a straight graph to reduce the relative effect of small reading differences. Divide distance values by time values, not vertical squares by horizontal squares unless the axis scales happen to make that ratio meaningful. Cumulative distance travelled cannot decrease; a falling line would need a different vertical quantity, such as signed position.
Pause and explain
What does the 4–6 s horizontal distance–time section mean?
Worked example
For (0,0), (4,12), (6,12), (10,20), find each straight-section speed.
Show the worked solution
- 0–4 s: (12 − 0)/(4 − 0) = 3 m/s.
- 4–6 s: no distance increase, so 0 m/s.
- 6–10 s: (20 − 12)/(10 − 6) = 2 m/s.
Answer 3 m/s; 0 m/s; 2 m/s.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Plot the given four points (0,0), (4,12), (6,12), (10,20) in s and m, with straight sections. Find the speeds.
Give me a hint
Calculate the gradient of each section separately.
Compare my reasoning
- Label time in s and distance in m.
- Find changes 12/4, 0/2 and 8/4.
- The section speeds are 3, 0 and 2 m/s.
3, 0 and 2 m/s.
Look for these in your work
- Label axes and plot correctly.
- Use differences within each section.
A straight distance–time section runs from (5 s, 18 m) to (9 s, 30 m). Find speed.
Give me a hint
Subtract both coordinates before dividing.
Compare my reasoning
- Distance change = 30 − 18 = 12 m.
- Time change = 9 − 5 = 4 s.
- Speed = 12/4 = 3 m/s.
3 m/s.
Look for these in your work
- Subtract the initial distance.
- Subtract the initial time.
A student says a straight rising distance–time line means acceleration. Explain the mistake.
Give me a hint
Compare gradients at different times on the straight line.
Compare my reasoning
- The gradient is the same everywhere on the section.
- Equal intervals of time add equal distances.
- The speed is constant; a changing speed needs a changing gradient.
A straight rising line has constant speed, not increasing speed.
Look for these in your work
- Link slope to speed.
- Distinguish distance increase from speed increase.
Common mistakes
- A horizontal distance line is rest, not constant non-zero speed.
- Use changes in axis values.
- A cumulative-distance graph cannot fall.
What can you explain now?
Distance rises from 12 m at 6 s to 20 m at 10 s. What is speed?
Compare with the explanation
2 m/s.
The gradient is 8 m over four seconds.
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Make it stick
Draw one journey with a stop and two different constant speeds, then label each gradient.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.