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# A car travelling at speed of 30 km/h is brought to a halt in 8 m by applying brakes. If the same car is travelling at 60 km/h, it can be brought to a halt with the same braking force in:A. 8 mB. 16 mC. 24 mD. 32 m

Last updated date: 20th Jun 2024
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Hint: Work done by the force is equal to change in kinetic energy of the body. Use the relation between work done and applied force. Also, take the final velocity of the car as zero.

Formula used:
$\dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 = Fd$
Here, ${v_f}$ is the final velocity of the car and ${v_i}$ is the initial velocity of the car, F is the braking force and d is the distance.

We know the relation between force and work done. The work done is the product of force and displacement of the body.
Therefore,
$W = Fd$ …… (1)
Here, F is the braking force and d is the distance travelled by the car after applying the brakes.
Also, the work done is the change in the kinetic energy of the object. Therefore, we can write,
$\dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 = W$ …… (2)
Here, ${v_f}$ is the final velocity of the car and ${v_i}$ is the initial velocity of the car.
Equate equations (1) and (2).
$\dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 = Fd$
Since the force is applied due to the brakes, the direction of the force is in the opposite direction of the motion of the car. Therefore, the above equation becomes,
$\dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 = - Fd$
Since the final velocity of the car is zero, the above equation becomes,
$\dfrac{1}{2}mv_i^2 = Fd$ …… (3)
For the car moving with velocity 60 km/h, we can write the above equation as follows,
$\dfrac{1}{2}mV_i^2 = Fd'$ …… (4)
Divide equation (2) by equation (1).
$\dfrac{{\dfrac{1}{2}mV_i^2}}{{\dfrac{1}{2}mv_i^2}} = \dfrac{{Fd'}}{{Fd}}$
$\Rightarrow \dfrac{{V_i^2}}{{v_i^2}} = \dfrac{{d'}}{d}$
$\therefore d' = {\left( {\dfrac{{{V_i}}}{{{v_i}}}} \right)^2}d$
Substitute 60 km/h for ${V_i}$, 30 km/h for ${v_i}$ and 8 m for d in the above equation.
$d' = {\left( {\dfrac{{60\,km/h}}{{30\,km/h}}} \right)^2}\left( {8\,m} \right)$
$\Rightarrow d' = 4 \times \left( {8\,m} \right)$
$\therefore d' = 32\,m$

So, the correct answer is “Option D”.

Note:
Since the braking force is the same in both the cases, you can also solve this question using kinematic relation ${v^2} = {u^2} + 2as$. Substitute 0 for final velocity of the car for both the cases and rearrange