Question

A truck and a car moving with the same kinetic energy are brought to rest by the application of brakes which provides equal retarding forces. Which of them will come to rest in a shorter distance?

• A. The truck
• B. The car
• C. Both will travel the same distance before coming to rest
• D. Cannot be predicted

Answer 1

The correct option is C

Both will travel the same distance before coming to rest

Step 1: Given

Kinetic Energy is same for both.

Retarding force is same for both.

Final velocity $v=0m{s}^{-1}$

Step 2: Formula used

Kinetic energy $K.E.=\frac{1}{2}m{v}^2$

Force, $F=ma$

$v^2=u^2+2as$

Step 3: Calculation of distance

Since retarding force is same

$$\begin{gathered} \Rightarrow F=-ma \\ \Rightarrow a=-\frac{F}{m}------\left(1\right) \end{gathered}$$

$$\begin{gathered} v^2=u^2+2as \\ 0=u^2+2as \\ {u}^2=-2as \end{gathered}$$

Substituting the value of $a$from eq (1)

$$\begin{gathered} u^2=-(-\frac{2Fs}{m}) \\ s=\left(\frac{1}{2}m{u}^2\right)\left(\frac{1}{F}\right) \\ s=\left(\frac{K.E.}{F}\right) \end{gathered}$$

Since, kinetic energy and force are the same for both car and truck, both will cover the same distance to come to rest.

Hence, option(c) is correct.