Q 11-02-129JEE MainJEE Main 2018 (15 Apr, Shift 1)Easy
An automobile, traveling at $40\ \text{km h}^{-1}$, can be stopped at a distance of $40\ \text{m}$ by applying brakes. If the same automobile is traveling at $80\ \text{km h}^{-1}$, the minimum stopping distance in metres is (Assume no skidding):
Answer: (D) $160\ \text{m}$
With the same maximum braking retardation $a$, the stopping distance is
$$s = \frac{u^2}{2a} \quad\Rightarrow\quad s \propto u^2$$
Doubling the speed makes the stopping distance four times:
$$s_2 = 40 \times \left(\frac{80}{40}\right)^2 = 160\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics