Question pattern: Velocity/acceleration–time graph analysis
Original question 2 of 2 in this pattern
Change in velocity from acceleration data
An object initially moves in the positive direction at 4.0 m s⁻¹. Its acceleration is constant within each interval shown.
a) Calculate the total change in velocity from t = 0 s to t = 9 s.
b) Determine the maximum speed during the interval.
c) Determine the velocity at t = 9 s and state whether the object reverses direction before or at that instant.
| Time interval / s | Acceleration / m s⁻² |
|---|---|
| 0–3 | +2.0 |
| 3–7 | 0 |
| 7–9 | −5.0 |
Worked answer
Solution
- Step 1
From 0–3 s:
Δv₁ = 2.0 × 3 = +6.0 m s⁻¹, so the speed reaches4.0 + 6.0 = 10.0 m s⁻¹. - Step 2
From 3–7 s:
Δv₂ = 0 × 4 = 0, so the maximum speed remains 10.0 m s⁻¹. - Step 3
From 7–9 s:
Δv₃ = -5.0 × 2 = -10.0 m s⁻¹. - Step 4
Total change:
+6.0 + 0 - 10.0 = -4.0 m s⁻¹; final velocity:4.0 - 4.0 = 0 m s⁻¹. The object reaches rest exactly at 9 s and has not yet reversed.
Final answer
Total Δv = −4.0 m s⁻¹; maximum speed = 10.0 m s⁻¹; final velocity = 0 m s⁻¹; no reversal occurs within the stated interval.
Common mistake
Claiming that the object reverses merely because the acceleration is negative.
Concepts tested
- acceleration–time graph
- change in velocity
- maximum speed
- direction reversal
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