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

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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 the initial velocity,$a$ is the retardation,and $s$ is the stopping distance.Since $v = 0$,we have $-u^2 = 2as$,which implies $s = \frac{u^2}{2|a|}$.This shows that the stopping distance $s \propto u^2$.Given $u_1 = 50 \,km/hr$ and $s_1 = 6 \,m$.For $u_2 = 100 \,km/hr$,the ratio is $\frac{s_2}{s_1} = \left(\frac{u_2}{u_1}ight)^2$.$\frac{s_2}{6} = \left(\frac{100}{50}ight)^2 = (2)^2 = 4$.Therefore,$s_2 = 4 	imes 6 = 24 \,m$.

$A$ car,moving with a speed of $50 \,km/hr$,can be stopped by brakes after at least $6 \,m$. If the same car is moving at a speed of $100 \,km/hr$,the minimum stopping distance is..........$m$

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