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If µ is the coefficient of friction between the tyres and road, then the minimum stopping distance for a car of mass m moving with velocity v is

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$\frac{v^{2}}{2 μg}$

$2 {μgv}^{2}$

$μg$

$μg + v^{2}$

answer is A.

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Detailed Solution

$(μmg) s = \frac{1}{2} {mv}^{2} or s = \frac{v^{2}}{2 μg}$

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