Question pattern: Motion graph construction & translation
Original question 2 of 2 in this pattern
Translate a velocity graph into position and acceleration
A velocity–time graph is made from straight-line segments joining the listed points.
a) State the acceleration in each time interval and use these values to sketch the corresponding acceleration–time graph.
b) Describe the shape of the position–time graph in each interval. State whether position ever decreases.
c) Calculate the total displacement from t = 0 s to t = 9 s.
| Time / s | Velocity / m s⁻¹ |
|---|---|
| 0 | +2 |
| 3 | +8 |
| 7 | +8 |
| 9 | 0 |
Worked answer
Solution
- Step 1
From 0–3 s:
a = (8 - 2) / 3 = +2.0 m s⁻². - Step 2
From 3–7 s: velocity is constant, so
a = 0. - Step 3
From 7–9 s:
a = (0 - 8) / 2 = -4.0 m s⁻². - Step 4
The position–time graph rises with increasing slope from 0–3 s, is a straight rising line from 3–7 s, and still rises but with decreasing slope from 7–9 s.
- Step 5
Position never decreases because velocity is never negative; it becomes horizontal only at the final instant.
- Step 6
Displacement is the v–t area:
1/2(2 + 8)3 + 8(4) + 1/2(8 + 0)2 = 15 + 32 + 8 = 55 m.
Final answer
Accelerations: +2.0, 0 and −4.0 m s⁻²; position increases throughout and levels off at t = 9 s; displacement = 55 m.
Common mistake
Drawing a decreasing position graph during the final interval simply because acceleration is negative.
Concepts tested
- graph translation
- acceleration–time graph
- position–time graph
- velocity–time area
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