Question pattern: Velocity/acceleration–time graph analysis
Original question 1 of 2 in this pattern
Signed area on a velocity–time graph
A velocity–time graph consists of straight-line segments joining the points in the table.
a) Determine the acceleration from t = 0 s to t = 4 s.
b) Calculate the displacement from t = 0 s to t = 11 s.
c) Calculate the total distance travelled over the same interval.
d) Explain why the answers to b) and c) are different.
| Time / s | Velocity / m s⁻¹ |
|---|---|
| 0 | 0 |
| 4 | +8 |
| 7 | +8 |
| 11 | −4 |
Worked answer
Solution
- Step 1
Initial acceleration:
a = (8 - 0) / 4 = 2.0 m s⁻². - Step 2
Area from 0–4 s:
1/2 × 4 × 8 = 16 m; area from 4–7 s:3 × 8 = 24 m. - Step 3
From 7–11 s the trapezium has signed area
1/2 × (8 + (-4)) × 4 = 8 m. Total displacement:16 + 24 + 8 = 48 m. - Step 4
The final segment has gradient
(-4 - 8) / 4 = -3 m s⁻², so velocity reaches zero after8/3 s, att = 9.67 s. - Step 5
Positive area from 7–9.67 s is
1/2 × (8/3) × 8 = 10.67 m; negative-region magnitude is1/2 × (4/3) × 4 = 2.67 m. - Step 6
Total distance:
16 + 24 + 10.67 + 2.67 = 53.3 m. Distance counts both directions positively, while displacement retains the negative area after reversal.
Final answer
Acceleration = 2.0 m s⁻²; displacement = 48 m; total distance = 53.3 m.
Common mistake
Using the signed trapezium area as the distance during the final segment.
Concepts tested
- velocity–time graph
- acceleration
- signed area
- distance
- displacement
- reversal
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