Question

A truck and a car are moving on a smooth, level road such that the kinetic energy associated with them is same. Brakes are applied simultaneously in both of them such that equal retarding forces are produced in them. Which one will cover a greater distance before it stops?

• A. Car
• B. Truck
• C. Both will cover the same distance
• D. Nothing can be decided

Answer 1

The correct option is C

Both will cover the same distance

Explanation for the correct option-

Option (C)

1. Let the Kinetic energy of truck is $K.E_T$, let the Kinetic energy of car is $K.E_C$.
2. Now, according to the question kinetic energy of both truck and car are same $K.E_T=K.E_C$
3. Here, we put the formula of kinetic energy that is $K.E=\frac{1}{2}m{v}^2(m=massandv=velocity)$
4. If $$\begin{gathered} m_1=massofthetruck{v}_1=velocityoftruck \\ {m}_2=massofthecar{v}_2=velocityofcar \end{gathered}$$
5. Then, $$\begin{gathered} \frac{1}{2}{m}_1{v_1}^2=\frac{1}{2}{m}_2{v_2}^2 \\ {m}_1{v_1}^2=m_2{v_2}^2 \end{gathered}$$
6. According to the third law of Newton-

For truck-

$$\begin{gathered} v^2=u^2+2aS(v=velocity,u=initialvelocity,a=acceleration,s=dis\tan cemoved) \\ 0={v_1}^2+2(-a)S_1 \\ {v_1}^2-2a{S}_1=0 \\ {S}_1=\frac{{v_1}^2}{2a} \end{gathered}$$

6. As, we know

$$\begin{gathered} f=ma(f=force) \\ a=\frac{f}{m_1}(fortruck) \end{gathered}$$

So, $S_1=\frac{v_1^2}{2{\displaystyle \frac{f}{m_1}}}=\frac{m_1{v}_1^2}{2f}$

7. For car same as above mentioned-

$S_2=\frac{{m_2{v}_2}^2}{2f}$

8. After comparing speed of both truck and car- $S_1=S_2(because{m}_1{v}_1^2=m_2{v}_2^2)$ and $f$ is same for both.

Thus, both will cover the same distance.

Hence, Option (C) is correct.