A constant power P is applied to a particle of mass m. The distance travelled by the particle when its velocity increases from v₁ to v₂ is (neglect friction )

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

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

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

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

answer is C.

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Detailed Solution

$\begin{array}{l} P = F v = m a v \Rightarrow a = \frac{P}{m v} \\ \Rightarrow v \frac{d v}{d s} = \frac{p}{m v} \Rightarrow v^{2} d v = \frac{p}{m} d s \\ \Rightarrow \frac{P}{m} \int_{0}^{s} d s = \int_{v_{1}}^{v_{2}} v^{2} d v \end{array}$

Integrate both sides under limits v from v₁ to v₂ds from 0 to S
We get $(\frac{v^{3}}{3}) = \frac{P}{m \cdot s}$
$\begin{aligned} \frac{1}{3} ({v_{2}}^{3} - v_{1}^{3}) = \frac{P}{m \cdot s} \\ \frac{M}{3} P (v_{2}^{3} - v_{1}^{3}) = S \end{aligned}$