Find distance from areas under speed-time graphs
Split a journey into rectangles and triangles and interpret their areas as travelled distance.
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
- Find areas of rectangles and triangles.
- Read durations by subtracting times.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What unit results from (m/s) × s?
What you will learn
- Calculate speed–time area for constant-speed and constant-acceleration sections.
- Combine sections without double-counting.
Area combines speed with duration
For constant speed, distance equals speed times time. A horizontal speed of 8 m/s for three seconds makes a rectangle of area 8 × 3 = 24 m. The graph units multiply as (m/s) × s = m; this explains why the area is a distance rather than a speed.
When speed rises uniformly from zero to 8 m/s over four seconds, the region under the straight line is a triangle. Its area is ½ × 4 × 8 = 16 m. A uniformly slowing section from 8 m/s to zero in two seconds similarly gives 8 m. Add the non-overlapping areas to obtain total travelled distance.
A non-zero start gives a trapezium
If speed changes uniformly from 4 to 10 m/s over six seconds, the area is a rectangle of 4 × 6 plus a triangle of ½ × 6 × 6. The total is 42 m. Equivalently use ½(4 + 10) × 6. The half-sum uses endpoint speeds only because the connecting section is straight.
For a curved speed graph, a single endpoint trapezium is generally an approximation, not an exact area. For the syllabus constant-speed or constant-acceleration problems here, use the specified straight sections. Time bases are durations: the interval seven to nine seconds lasts two seconds, not nine.
Pause and explain
For a uniform speed rise from 4 to 10 m/s over six seconds, what is distance?
Worked example
Find distance for the graph through (0,0), (4,8), (7,8), (9,0).
Show the worked solution
- 0–4 s triangle = ½ × 4 × 8 = 16 m.
- 4–7 s rectangle = 3 × 8 = 24 m.
- 7–9 s triangle = ½ × 2 × 8 = 8 m; total = 48 m.
Answer 48 m.
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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.
Find total distance for straight graph sections (0,0), (4,8), (7,8), (9,0) in seconds and m/s.
Give me a hint
Add two triangles and the middle rectangle.
Compare my reasoning
- First triangle gives 16 m.
- Middle rectangle gives 24 m.
- Final triangle gives 8 m; sum is 48 m.
48 m.
Look for these in your work
- Use interval durations.
- Add all non-overlapping areas.
Speed increases uniformly from 4 to 10 m/s over six seconds. Find distance.
Give me a hint
A non-zero initial speed creates a trapezium.
Compare my reasoning
- Rectangle area = 4 × 6 = 24 m.
- Extra triangle = ½ × 6 × 6 = 18 m.
- Total distance = 42 m.
42 m.
Look for these in your work
- Include the base rectangle.
- Attach distance units.
For the nine-second 48 m journey in the diagram, calculate whole-journey average speed and explain whether it was constant.
Give me a hint
Use area for total distance, then divide by total time.
Compare my reasoning
- Total distance from the graph is 48 m.
- 48/9 ≈ 5.33 m/s.
- The graph rises, stays level and falls, so this average was not the speed at every instant.
Approximately 5.33 m/s; the speed was not constant.
Look for these in your work
- Use the total graph area.
- Distinguish average from instantaneous speed.
Common mistakes
- The time base is an interval difference.
- A non-zero starting speed needs its rectangle as well.
- Endpoint averaging is exact here only for a straight speed section.
What can you explain now?
Speed increases uniformly from 4 to 10 m/s in six seconds. What distance is travelled?
Compare with the explanation
42 m.
The trapezium area is ½(4 + 10) × 6.
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Make it stick
Split a new straight-section speed graph into areas and check the resulting distance unit.
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.