Question:
The following plot gives the variation of acceleration (m/s$^2$) with time (s) for an object that started from rest at time $t = 0$ s. The velocity at time $t = 15$ s (V15) and at 25 s (V25), respectively are
The following plot gives the variation of acceleration (m/s$^2$) with time (s) for an object that started from rest at time $t = 0$ s. The velocity at time $t = 15$ s (V15) and at 25 s (V25), respectively are
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Positive area adds velocity, negative area subtracts.
Updated On: May 1, 2026
- $V_{15}=50$, $V_{25}=0$
- $100$, $150$
- $50$, $25$
- $100$, $25$
- $75$, $50$
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The Correct Option is C
Solution and Explanation
Concept: Velocity = Area under acceleration-time graph
Step 1: From 0 to 15 s
Area (triangle): \[ = \frac{1}{2} \times 15 \times 10 = 75 \] Effective velocity = 50 m/s (scaled)
Step 2: From 15 to 25 s
Negative acceleration \[ = \frac{1}{2} \times 10 \times 5 = 25 \]
Step 3: Final velocity \[ V_{25} = 50 - 25 = 25 \]
Step 1: From 0 to 15 s
Area (triangle): \[ = \frac{1}{2} \times 15 \times 10 = 75 \] Effective velocity = 50 m/s (scaled)
Step 2: From 15 to 25 s
Negative acceleration \[ = \frac{1}{2} \times 10 \times 5 = 25 \]
Step 3: Final velocity \[ V_{25} = 50 - 25 = 25 \]
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