Explain why the area under a velocity–time graph between two instants gives the displacement of an object during that interval, even when the velocity is not constant.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
For any small time interval δt the displacement is approximately v·δt, which is the area of a narrow rectangular strip under the graph.
The total displacement is found by summing all such strips across the entire interval (integration / summation of strips).
As δt approaches zero the approximation becomes exact, so the sum of all strip areas equals the exact area under the curve.
Regions of the graph below the time axis (negative velocity) contribute negative area, correctly representing motion in the opposite direction, so the net area gives displacement rather than distance.
The total displacement is found by summing all such strips across the entire interval (integration / summation of strips).
As δt approaches zero the approximation becomes exact, so the sum of all strip areas equals the exact area under the curve.
Regions of the graph below the time axis (negative velocity) contribute negative area, correctly representing motion in the opposite direction, so the net area gives displacement rather than distance.
Examiner tips
- Use the phrase ‘area under the curve’ to link to displacement
- Show the integral form ∫v dt if possible
- Mention negative areas for opposite direction
Common mistakes
- Confusing displacement with distance – ignoring negative areas
- Failing to explain the limit δt→0 or integration concept
Mark scheme (4 marks)
- For any small time interval δt, the displacement is approximately equal to v·δt, which is the area of a narrow rectangular strip under the graph.
- The total displacement is found by summing all such strips across the entire interval (integration / summation of strips).
- As δt approaches zero the approximation becomes exact, so the sum of all strip areas equals the exact area under the curve.
- Regions of the graph below the time axis (negative velocity) contribute negative area, correctly representing motion in the opposite direction, so the net area gives displacement rather than distance.
Key terms in this question
displacement · velocity–time graph
Related
- All IB DP Physics Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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