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Q.

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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a

$\frac{v_{0}}{μ g}$

b

${(\frac{v_{0}}{μ g})}^{2}$

c

$\frac{v_{0}}{μ}$

d

$\frac{v_{0}^{2}}{2 μ g}$

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, $v^{2} = u^{2} - 2 a s$

$∴ 0 = u^{2} - 2 a s$ Or $s = \frac{u^{2}}{2 a} = \frac{u^{2}}{2 μ g} = \frac{v_{0}^{2}}{2 μ g}$

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A car is moving along a straight horizontal road with a speed v0. If the coefficient of friction between the tyres and the road is μ , the shortest distance in which the car can be stopped is