Question

A car travelling at a speed of 30 km/h is brought to a halt in 8m 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 m
(B) 16 m
(C) 24 m
(D) 32 m

Answer 1

Hint: Coming to a halt signifies that the velocity was decelerated to zero. Use the third equation of motion to calculate the deceleration of the body.

Formula used: In this solution we will be using the following formulae;
 $ {v^2} = {u^2} - 2as $ where $ v $ is the final velocity of a body, $ u $ is the initial velocity of the body, $ a $ is the magnitude of the deceleration, and $ s $ is the distance travelled as it decelerates from final velocity to initial velocity.

Complete step by step answer:
To solve the above question, we find the deceleration of the car when the brake was applied.
This can be found using the third equation of motion, which is
 $ {v^2} = {u^2} - 2as $ where $ v $ is the final velocity of a body, $ u $ is the initial velocity of the body, $ a $ is the magnitude of the deceleration, and $ s $ is the distance travelled as it decelerates from final velocity to initial velocity.
Hence, inserting known formula, we get
 $ 0 = {8.33^2} - 2a\left( 8 \right) $ since 30 km/h is $ 8.33m/s $
Hence,
 $ a = \dfrac{{{{8.33}^2}}}{{2\left( 8 \right)}} = 4.34m/{s^2} $
This signifies that for this particular car, the brake force will cause an acceleration of the above value.
Hence, the distance it would travel if the initial velocity was 60 km/h would be
 $ 0 = {16.67^2} - 2\left( {4.34} \right)s $
Hence, by calculating for $ s $ , we have
 $ s = \dfrac{{{{16.67}^2}}}{{2\left( {4.34} \right)}} = 32m $
Thus, the correct option is D.

Note:
Alternatively, if we note from the equation of motion, that the since the final velocity is zero, then the initial velocity is proportional to the square root of the distance as in
 $ {u^2} = 2as $
 $ \Rightarrow u = \sqrt {2a} \sqrt s = k\sqrt s $
Hence, we can simply right that
 $ \dfrac{{{u_1}}}{{\sqrt {{s_1}} }} = \dfrac{{{u_2}}}{{\sqrt {{s_2}} }} $
Hence, making $ \sqrt {{s_2}} $ subject, we get
 $ \sqrt {{s_2}} = \dfrac{{\sqrt {{s_1}} }}{{{u_1}}}{u_2} $
We can insert the values without converting to SI since they cancel out
 $ \sqrt {{s_2}} = \dfrac{{\sqrt 8 }}{{30}} \times 60 $
By calculation, we have
 $ \sqrt {{s_2}} = 2\sqrt 8 $
Hence, squaring both sides, we get
 $ {s_2} = 4 \times 8 = 32m $ .