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} $



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