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A constant power P is applied to a particle of mass m. The distance travelled by the particle when its velocity increases from $v_{1} to v_{2}$ is (neglect friction) :

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

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

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

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

answer is A.

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Detailed Solution

P = Fv = mav $\Rightarrow a = \frac{P}{mv}$

$\Rightarrow v \frac{dv}{ds} = \frac{P}{mv} \Rightarrow v^{2} dv = \frac{P}{m} ds$

$\Rightarrow \frac{P}{m} \int_{0}^{s} ds = \int_{v_{1}}^{v_{2}} v^{2} dv \Rightarrow \frac{P}{m} s = \frac{1}{3} ({v_{2}}^{3} - {v_{1}}^{3})$

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

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