Laws Of Motion Question 99
Question: A car is moving along a straight horizontal road with a speed $ v _{0} $ . If the coefficient of friction between the tyres and the road is $ \mu $ , the shortest distance in which the car can be stopped is[MP PET 1985; BHU 2002]
Options:
A)$ \frac{v _{0}^{2}}{2\mu g} $
B) $ \frac{v _{0}}{\mu g} $
C) $ {{( \frac{v _{0}}{\mu g} )}^{2}} $
D) $ \frac{v _{0}}{\mu } $
Show Answer
Answer:
Correct Answer: A
Solution:
Retarding force $ F=ma=\mu R=\mu \ mg $
$ \therefore $ $ a=\mu g $ Now from equation of motion$ v^{2}=u^{2}-2as $
$ \Rightarrow \ 0=u^{2}-2as $
therefore $ s=\frac{u^{2}}{2a}=\frac{u^{2}}{2\mu \ g} $
$ \therefore $ $ =\frac{v _{0}^{2}}{2\mu g} $
BETA
