A car is moving along a straight horizontal road with a speed $u_{0}$ . 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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$\frac{u_{0}}{μg}$

${(\frac{u_{0}}{μg})}^{2}$

$\frac{u_{0}}{μ}$

$\frac{u_{\circ}^{2}}{2 μg}$

answer is A.

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Detailed Solution

Retarding force $F = μR = μmg$
$∴ a = μg$
Now from equation of motion $v^{2} = u^{2} - 2 as$
$\begin{array}{l} \Rightarrow 0 = u^{2} - 2 as \Rightarrow s = \frac{u^{2}}{2 a} = \frac{u^{2}}{2 μg} \\ ∴ s = \frac{u_{0}^{2}}{2 μg} \end{array}$

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