A car moving at a speed of 50 , km/hr stops a | vedclass

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(D) Using the equation of motion $v^2 = u^2 + 2as$,where $v = 0$ (final velocity),$u$ is initial velocity,$a$ is retardation,and $s$ is the stopping d

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$24$

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(D) Using the equation of motion $v^2 = u^2 + 2as$,where $v = 0$ (final velocity),$u$ is initial velocity,$a$ is retardation,and $s$ is the stopping distance.Since $0 = u^2 - 2as$,we get $s = \frac{u^2}{2a}$.Assuming the retardation $a$ is constant,the stopping distance $s$ is directly proportional to the square of the initial velocity: $s \propto u^2$.Therefore,$\frac{s_2}{s_1} = \left( \frac{u_2}{u_1} ight)^2$.Given $u_1 = 50 \, km/hr$,$s_1 = 6 \, m$,and $u_2 = 100 \, km/hr$.$\frac{s_2}{6} = \left( \frac{100}{50} ight)^2 = (2)^2 = 4$.$s_2 = 4 	imes 6 = 24 \, m$.

$A$ car moving at a speed of $50 \, km/hr$ stops after covering a distance of $6 \, m$ upon applying brakes. If the same car is moving at a speed of $100 \, km/hr$,what distance (in $m$) will it cover after applying the brakes?

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