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(A) The retarding force $F$ acting on the car is due to friction,given by $F = \mu R = \mu mg$.According to Newton's second law,$F = ma$,so $ma = \mu

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$\frac{v_0^2}{2\mu g}$

Vedclass · Answer

(A) The retarding force $F$ acting on the car is due to friction,given by $F = \mu R = \mu mg$.According to Newton's second law,$F = ma$,so $ma = \mu mg$,which gives the retardation $a = \mu g$.Using the third equation of motion,$v^2 = u^2 - 2as$,where $v$ is the final velocity $(0)$,$u$ is the initial velocity $(v_0)$,and $s$ is the stopping distance.Substituting the values: $0^2 = v_0^2 - 2(\mu g)s$.Rearranging for $s$,we get $2\mu gs = v_0^2$.Therefore,the shortest stopping distance is $s = \frac{v_0^2}{2\mu g}$.

$A$ 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

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