Q. Consider a car moving along a straight horizontal road with a speed of$\ 72\ km/h$. If the coefficient kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is $\lbrack g = 10\ ms^{- 2}\rbrack$
A) $30\ m$
B) $40\ m$
C) $72\ m$
D) $20\ m$
Explanation
(b) $s = \frac{u^{2}}{2\mu g} = \frac{(20)^{2}}{2 \times 0.5 \times 10} = 40\ m$,
$72\ km/hr = 72 \times \frac{5}{18} = 20\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.