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(B) The aerodynamic drag force $F$ is proportional to speed $v$,so $F = kv$,where $k$ is a constant.The power $P$ delivered by the engine to overcome

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

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(B) The aerodynamic drag force $F$ is proportional to speed $v$,so $F = kv$,where $k$ is a constant.The power $P$ delivered by the engine to overcome this drag force at constant speed is given by $P = F \cdot v$.Substituting $F = kv$ into the power equation,we get $P = (kv) \cdot v = kv^2$.From this,we can express the speed $v$ as $v = \sqrt{\frac{P}{k}}$.If the power output is doubled,the new power $P' = 2P$.The new maximum speed $v'$ will be $v' = \sqrt{\frac{P'}{k}} = \sqrt{\frac{2P}{k}} = \sqrt{2} \cdot \sqrt{\frac{P}{k}}$.Since $v = \sqrt{\frac{P}{k}}$,we have $v' = \sqrt{2}v$.Therefore,the maximum speed increases by a factor of $\sqrt{2}$.

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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