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(B) The aerodynamic drag force $F$ is proportional to the speed $V$,so $F = kV$,where $k$ is a constant.Power $P$ is defined as the product of force a

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increases by a factor of $\sqrt{2}$

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(B) The aerodynamic drag force $F$ is proportional to the speed $V$,so $F = kV$,where $k$ is a constant.Power $P$ is defined as the product of force and velocity: $P = F \cdot V$.Substituting $F = kV$ into the power equation,we get $P = (kV) \cdot V = kV^2$.This implies that $P \propto V^2$,or $V \propto \sqrt{P}$.If the power output $P$ is doubled to $2P$,the new speed $V'$ becomes $V' \propto \sqrt{2P} = \sqrt{2} \cdot \sqrt{P}$.Therefore,the new speed $V'$ is $\sqrt{2}$ times the original speed $V$.

Assume the aerodynamic drag force on a car is proportional to its speed. If the power output from the engine is doubled,then the maximum speed of the car:

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