# 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.

> IB DP Physics Higher Level (2023 syllabus) — A.1 Kinematics · Explain · 4 marks

## Mark scheme (4 marks)

1. 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.
2. The total displacement is found by summing all such strips across the entire interval (integration / summation of strips).
3. As δt approaches zero the approximation becomes exact, so the sum of all strip areas equals the exact area under the curve.
4. 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

- [displacement](https://www.gradenine.co.uk/glossary/displacement)
- [velocity–time graph](https://www.gradenine.co.uk/glossary/velocity-time-graph)

## Related

- [Revision notes for IB DP Physics Higher Level (2023 syllabus)](https://www.gradenine.co.uk/learn)
- [How to answer "Explain" questions](https://www.gradenine.co.uk/tools/command-word-cheatsheet)
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Source: [GradeNine](https://www.gradenine.co.uk/q/explain-why-the-area-under-a-62f4fcc7) · Published by Druglandscape Ltd.