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(B) We know that power $P = Fv = (ma)v$.Given $a = \frac{P}{mv}$,we can write acceleration as $a = v \frac{dv}{ds}$.Substituting this into the power e

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$\frac{m}{3P}(v_2^3 - v_1^3)$

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(B) We know that power $P = Fv = (ma)v$.Given $a = \frac{P}{mv}$,we can write acceleration as $a = v \frac{dv}{ds}$.Substituting this into the power equation: $P = m(v \frac{dv}{ds})v = m v^2 \frac{dv}{ds}$.Rearranging the terms to integrate: $ds = \frac{m}{P} v^2 dv$.Integrating both sides from initial velocity $v_1$ to final velocity $v_2$ and distance $0$ to $s$:$\int_{0}^{s} ds = \frac{m}{P} \int_{v_1}^{v_2} v^2 dv$.$s = \frac{m}{P} [\frac{v^3}{3}]_{v_1}^{v_2}$.$s = \frac{m}{3P} (v_2^3 - v_1^3)$.

An automobile engine of mass $m$ provides constant power $P$. The acceleration is given by $a = \frac{P}{mv}$. (Assume friction is absent). What is the distance covered by the vehicle while its velocity increases from $V_1$ to $V_2$?

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