A car moving with a velocity of 20 ,m ,s^{-1} | vedclass

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

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$160$

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(D) Using the third equation of motion, $v^2 - u^2 = 2as$, where $v = 0$ (final velocity), $u$ is initial velocity, $a$ is retardation $(-a)$, and $s$ is distance.$0^2 - u^2 = 2(-a)s \Rightarrow u^2 = 2as \Rightarrow s = \frac{u^2}{2a}$.Since the retardation $a$ is constant, $s \propto u^2$.For the first case: $40 = \frac{(20)^2}{2a} \Rightarrow 2a = \frac{400}{40} = 10 \,m \,s^{-2}$.For the second case, the new velocity $u' = 2 	imes 20 = 40 \,m \,s^{-1}$.The new distance $s' = \frac{(u')^2}{2a} = \frac{(40)^2}{10} = \frac{1600}{10} = 160 \,m$.

$A$ car moving with a velocity of $20 \,m \,s^{-1}$ is stopped in a distance of $40 \,m$. If the same car is travelling at double the velocity, the distance travelled by it for the same retardation is (in $\,m$)

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