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Physics-

General

Easy

Question

A car is moving along a straight horizontal road with a speed $v subscript 0 end subscript$ . If the coefficient of friction between tyres and the road is $mu$ , the shortest distance in which the car can be stopped is

$\frac{v_{0}^{2}}{2 μ g}$
$\frac{v_{0}}{μ g}$
${(\frac{v_{0}}{μ g})}^{2}$
$\frac{v_{0}}{μ}$

The correct answer is: $fraction numerator v subscript 0 end subscript superscript 2 end superscript over denominator 2 mu g end fraction$

Retarding force $F equals m a equals mu R equals mu blank m g$
$a equals mu g$
Now, from equation of motion, $v to the power of 2 end exponent equals u to the power of 2 end exponent minus 2 a s$
$therefore blank 0 equals u to the power of 2 end exponent minus 2 a s$
$therefore blank s equals fraction numerator u to the power of 2 end exponent over denominator 2 a end fraction equals fraction numerator u to the power of 2 end exponent over denominator 2 mu g end fraction equals fraction numerator v subscript 0 end subscript superscript 2 end superscript over denominator 2 mu g end fraction$

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