A car travelling along a straight road accelerates uniformly from rest. Explain how the velocity–time graph for this motion differs from the displacement–time graph for the same motion, and what each graph reveals about the car's acceleration.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
A straight line through the origin – the slope is the constant acceleration.
A curved (parabolic) s‑t graph – the slope (velocity) increases with time.
The slope of the v‑t graph gives the uniform acceleration.
The increasing slope of the s‑t graph shows the car is accelerating (its second derivative is the acceleration).
A curved (parabolic) s‑t graph – the slope (velocity) increases with time.
The slope of the v‑t graph gives the uniform acceleration.
The increasing slope of the s‑t graph shows the car is accelerating (its second derivative is the acceleration).
Examiner tips
- Use the word ‘slope’ to link gradient with acceleration.
- Show the v‑t graph is linear and the s‑t graph is a parabola.
- Mention that the v‑t slope is constant, the s‑t slope is increasing.
- Keep the answer concise – 4 points only.
Mark scheme (4 marks)
- The velocity–time graph is a straight line (through the origin) with a positive gradient
- The displacement–time graph is a curve (parabola / increasing gradient) starting from the origin
- The gradient of the velocity–time graph gives / equals the (uniform) acceleration
- The gradient of the displacement–time graph increases with time, confirming the car is accelerating / non-zero acceleration; OR the acceleration equals the second derivative / rate of change of the gradient of the s–t graph
Key terms in this question
velocity–time graph · displacement–time graph
Related
- All IB DP Physics Standard Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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