Question:
If a particle is moving such that the velocity acquired is proportional to the square root of the distance covered, then its acceleration is
If a particle is moving such that the velocity acquired is proportional to the square root of the distance covered, then its acceleration is
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If a particle is moving such that the velocity acquired is proportional to the square root of the distance covered, then its acceleration is
Updated On: Apr 15, 2026
- a constant
- $\propto s^{2}$
- $\propto\frac{1}{s^{2}}$
- $\propto\frac{1}{s}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Given $v \propto \sqrt{s} \Rightarrow v = \lambda \sqrt{s}$.
Step 2: Analysis
Differentiate velocity with respect to distance: $\frac{dv}{ds} = \frac{\lambda}{2\sqrt{s}}$.
Step 3: Evaluation
Acceleration $a = v \frac{dv}{ds} = (\lambda \sqrt{s}) \cdot (\frac{\lambda}{2\sqrt{s}})$.
Step 4: Conclusion
$a = \frac{\lambda^2}{2}$, which is a constant.
Final Answer: (a)
Given $v \propto \sqrt{s} \Rightarrow v = \lambda \sqrt{s}$.
Step 2: Analysis
Differentiate velocity with respect to distance: $\frac{dv}{ds} = \frac{\lambda}{2\sqrt{s}}$.
Step 3: Evaluation
Acceleration $a = v \frac{dv}{ds} = (\lambda \sqrt{s}) \cdot (\frac{\lambda}{2\sqrt{s}})$.
Step 4: Conclusion
$a = \frac{\lambda^2}{2}$, which is a constant.
Final Answer: (a)
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