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(C) Power $P$ is defined as the rate of change of work done, $P = \frac{dW}{dt} = Fv = (ma)v = m \left(\frac{dv}{dt}\right) v$.Since $P$ is constant,

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$v^3 \propto \frac{2 P}{m}$

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(C) Power $P$ is defined as the rate of change of work done, $P = \frac{dW}{dt} = Fv = (ma)v = m \left(\frac{dv}{dt}\right) v$.Since $P$ is constant, we have $P dt = mv dv$.Integrating both sides from $t=0$ to $t$ and $v=0$ to $v$, we get $Pt = \frac{1}{2}mv^2$.Thus, $v^2 = \frac{2Pt}{m}$.Also, $v = \frac{dx}{dt}$, so $\frac{dx}{dt} = \sqrt{\frac{2P}{m}} t^{1/2}$.Integrating with respect to $t$, $x = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2}$, which implies $t^{3/2} \propto x \Rightarrow t \propto x^{2/3}$.Substituting $t$ back into the velocity equation: $v^2 \propto t \propto x^{2/3}$, so $v \propto x^{1/3}$.However, looking at the relationship between $v$ and $P$ at a fixed distance $x$, we have $v^3 \propto P$. Thus, $v^3 \propto \frac{P}{m}$.

The velocity $v$ reached by a car of mass $m$ at a certain distance from the starting point, driven with constant power $P$, is such that:

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