Find distance from areas under speed-time graphs
Split a journey into rectangles and triangles and interpret their areas as travelled distance.
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
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.
Sources for this lesson
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.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
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.