Q. An automobile travelling with a speed of$\ 60\ km/h$, can brake to stop within a distance of $20\ m$. If the car is going twice at fast$,\ i.e$. $120\ km/h$, the stopping distance will be
A) $20\ m$
B) $40\ m$
C) $60\ m$
D) $80\ m$
Explanation
(d) The stopping distance, $S \propto u^{2}\ \left( \because v^{2} = u^{2} - 2as \right)$
$\Rightarrow \frac{S_{2}}{S_{1}} = \left( \frac{u_{2}}{u_{1}} \right)^{2} = \left( \frac{120}{60} \right)^{2} = 4$
$\Rightarrow S_{2} = 4 \times S_{1} = 4 \times 20 = 80\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.