An automobile travelling with a speed of 60,k | vedclass

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(D) The stopping distance $S$ of a vehicle is given by the relation $v^2 = u^2 + 2as$. Since the final velocity $v = 0$,we have $0 = u^2 - 2a|S|$,whic

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(D) The stopping distance $S$ of a vehicle is given by the relation $v^2 = u^2 + 2as$. Since the final velocity $v = 0$,we have $0 = u^2 - 2a|S|$,which implies $S = \frac{u^2}{2|a|}$.Since the deceleration $a$ is constant,the stopping distance is directly proportional to the square of the initial velocity: $S \propto u^2$.Given $S_1 = 20\,m$ at $u_1 = 60\,km/h$ and $u_2 = 120\,km/h$.Using the ratio: $\frac{S_2}{S_1} = \left(\frac{u_2}{u_1}ight)^2 = \left(\frac{120}{60}ight)^2 = (2)^2 = 4$.Therefore,$S_2 = 4 	imes S_1 = 4 	imes 20\,m = 80\,m$.

An automobile travelling with a speed of $60\,km/h$ can brake to stop within a distance of $20\,m$. If the car is going twice as fast,i.e.,$120\,km/h$,the stopping distance will be ........... $m$.

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