Q 11-02-022NEETJEE MainTop questionEasy
A car moving at $54$ km/h can be stopped by its brakes in a distance of $15$ m. With the same braking retardation, the minimum stopping distance when it moves at $72$ km/h is
Answer: (C) $26.7$ m
With constant retardation $a$: $0 = u^2 - 2as$, so $s = \dfrac{u^2}{2a} \propto u^2$.
$$\frac{s_2}{s_1} = \left(\frac{72}{54}\right)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$$
$$s_2 = 15 \times \frac{16}{9} = 26.7\ \text{m}$$
A $33\%$ higher speed needs $78\%$ more stopping distance.
Solution by Sreeraj P, M.Sc Physics