A car is moving with velocity V. It stops aft | vedclass

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(C) Using the third equation of motion: $v^2 = u^2 - 2as$. Since the car stops,the final velocity $v = 0$. Therefore,$0 = u^2 - 2as$,which gives $s =

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) Using the third equation of motion: $v^2 = u^2 - 2as$. Since the car stops,the final velocity $v = 0$. Therefore,$0 = u^2 - 2as$,which gives $s = \frac{u^2}{2a}$. This implies that the stopping distance $s$ is directly proportional to the square of the initial velocity,i.e.,$s \propto u^2$. Let the initial velocity be $u_1 = V$ and the stopping distance be $s_1 = 20 \ m$. When the velocity is doubled,$u_2 = 2V$. Then,$\frac{s_2}{s_1} = \left( \frac{u_2}{u_1} ight)^2 = \left( \frac{2V}{V} ight)^2 = 4$. Thus,$s_2 = 4 	imes s_1 = 4 	imes 20 \ m = 80 \ m$.

$A$ car is moving with velocity $V$. It stops after applying brakes at a distance of $20 \ m$. If the velocity of the car is doubled,then what distance will it cover after applying the brakes (in $m$)?

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