Question

Assume the aerodynamic drag force on a car is proportional to its speed. If the power output from the engine is double, then the maximum speed of the car.
A. is unchanged
B. increases by a factor of \[\sqrt{2}\]
C. is also double
D. increases by a factor of four

Answer 1

Hint: This question can be solved using the formula that relates power, force and velocity. Firstly, we will compute the expression for the initial power and the initial speed and then, we will compute the expression for the output power and the final speed, by making use of the initial power and speed equations to find the maximum speed of the car.
Formula used:
\[P=\dfrac{W}{t}\]

Complete step by step answer:
The average power formula is given as follows.
\[P=\dfrac{W}{t}\]
Where W is the work done and t is the time taken.
This work done can be expressed as the product of force and the change in displacement. So, we get,
\[\begin{align}
  & P=\dfrac{F\times \Delta d}{\Delta t} \\
 & \Rightarrow P=F\dfrac{\Delta d}{\Delta t} \\
\end{align}\]
The second term in the RHS part of the above equation represents the velocity.
\[P=FV\]
We will make use of this formula to carry out the further calculation.
From the data, we have data that the aerodynamic drag force on a car is proportional to its speed.
Let ‘v’ denote the speed of the car.
\[\begin{align}
  & F\propto v \\
 & \Rightarrow F=kv \\
\end{align}\]
Where k is a proportionality constant.
Rearrange the terms of the above equation to represent the equation in terms of the speed.
\[v=\dfrac{F}{k}\]….. (1)
Now consider the given data, that is, the power output from the engine is double.
Let ‘P’ denote the initial power, using the power equation, we have,
\[P=Fv\]
Rearrange the terms of the above equation to represent the equation in terms of the force.
\[F=\dfrac{P}{v}\]….. (2)
Equate the equation (2) in (1).
\[\begin{align}
  & v=\dfrac{{}^{P}/{}_{v}}{k} \\
 & \Rightarrow v=\dfrac{P}{vk} \\
 & \Rightarrow {{v}^{2}}=\dfrac{P}{k} \\
\end{align}\]
Thus, the equation is,
\[v=\sqrt{\dfrac{P}{k}}\]
Let P’ denote the output power and from the given data, we have, power output from the engine is double. So, we get,
\[P'=2P\]
The maximum speed of the car is,
\[v'=\sqrt{\dfrac{P'}{k}}\]
Substitute the value of the initial power.
\[\begin{align}
  & v'=\sqrt{\dfrac{2P}{k}} \\
 & \Rightarrow v'=\sqrt{2}\sqrt{\dfrac{P}{k}} \\
 & \Rightarrow v'=\sqrt{2}v \\
\end{align}\]
The maximum power of a car increases by a factor of \[\sqrt{2}\].

Thus, the option (B) is correct.

Note:
This is a concept based on the relation between the power and the speed of the car. We have derived the equation that relates the power, force and the speed, as the condition in terms of the force was given. Then considering two situations and equating the equations, we have found the maximum speed of the car.