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A constant power ‘P’ is applied to a particle of mass ‘m’, the displacement of the particle when its velocity increases from $v_{1}$ to $v_{2}$ is ( ignore friction)

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$s = \frac{m}{3 P} (v_{2}^{3} - v_{1}^{3})$

$s = \frac{1}{2} \frac{m}{P} (v_{1}^{2} - v_{2}^{2})$

$s = \frac{m}{2 P} (v_{2}^{3} - v_{1}^{3})$

$s = \frac{m}{P} (v_{2}^{3} - v_{1}^{3})$

answer is A.

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Detailed Solution

P=F.V=ma(v)

$a = \frac{p}{m v} \Rightarrow v . \frac{d v}{d s} = \frac{p}{m v}$

$v^{2} d v = \frac{p}{m} d s$

Integrate both sides

$\frac{p}{m} \int d s = \int_{v_{1}}^{v_{2}} v^{2} d v$

$\frac{p}{m} s = \frac{v_{2}^{3} - v_{1}^{3}}{3}$

$s = \frac{m (v_{2}^{2} - v_{1}^{3})}{3 p}$

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