Question

A moving car encounters air resistance which is proportional to the square of the speed of the car. The ratio of the power required at $40km{h}^{-1}$ to that required at $80km{h}^{-1}$, with the same braking power, is

• A. $1:6$
• B. $1:8$
• C. $1:16$
• D. $16:1$

Answer 1

The correct option is B

$1:8$

Step 1: Given data

The initial speed of the car is, $v_1=40km{h}^{-1}$

The final speed of the car is, $v_2=80km{h}^{-1}$

Step 2: Finding the ratio of the power

Air resistance is proportional to the square of the speed of the car. $F$ is directly proportional to $v^2$ where $F$ is the air resistance (negligible) and $v$ is the speed of the car.

Power $P$ can be written as $P=Fv$.

The ratio of the powers can be calculated as,

$\begin{array}{rcl}\frac{P_1}{P_2}& =& \frac{{v_1}^3}{{v_2}^3}\\ & =& \frac{40km{h}^{-1}\times 40km{h}^{-1}\times 40km{h}^{-1}}{80km{h}^{-1}\times 80km{h}^{-1}\times 80km{h}^{-1}}\\ & =& \frac{1}{8}\mathrm{or}1:8\end{array}$

The ratio of the powers can be written as $1:8$

Hence, option (b) is correct.