Question:
A particle moves with velocity $v = at - bt^2$ where a and b are constants. The acceleration becomes zero at:
A particle moves with velocity $v = at - bt^2$ where a and b are constants. The acceleration becomes zero at:
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Differentiate velocity once to get acceleration, then solve for t when it equals zero.
Updated On: Jun 10, 2026
- $t = \frac{a}{b}$
- $t = \frac{a}{2b}$
- $t = \frac{2a}{b}$
- $t = \frac{b}{a}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Acceleration ($a_{acc}$) is the time derivative of velocity ($v$).
Step 2: Analysis
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(at - bt^2) = a - 2bt$. Set $a_{acc} = 0$ to find the time: $a - 2bt = 0 \implies t = \frac{a}{2b}$.
Step 3: Conclusion
The acceleration becomes zero at $t = \frac{a}{2b}$.
Final Answer: (B)
Acceleration ($a_{acc}$) is the time derivative of velocity ($v$).
Step 2: Analysis
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(at - bt^2) = a - 2bt$. Set $a_{acc} = 0$ to find the time: $a - 2bt = 0 \implies t = \frac{a}{2b}$.
Step 3: Conclusion
The acceleration becomes zero at $t = \frac{a}{2b}$.
Final Answer: (B)
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