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 $μ$ , the shortest distance in which the car can be stopped is

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$\frac{v_{0}}{μ g}$

$\frac{v_{0}^{2}}{2 μ g}$

$\frac{v_{0}}{μ}$

${(\frac{v_{0}}{μ g})}^{2}$

answer is A.

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Detailed Solution

Retarding force $F = m a = μ ⥄ R = μ m g$
$a = μ g$
Now from equation of motion
$v^{2} = u^{2} - 2 a s \Rightarrow 0 = u^{2} - 2 a s$
$\Rightarrow s = \frac{u^{2}}{2 a} = \frac{u^{2}}{2 μ g} ∴= \frac{v_{0}^{2}}{2 μ g}$

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