(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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(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 $u' = 2 \times 20 = 40 \,m \,s^{-1}$.The new distance $s' = \frac{(u')^2}{2a} = \frac{(40)^2}{10} = \frac{1600}{10} = 160 \,m$.

Class 11 Physics Motion in a Straight Line Uniformly Accelerated Motion

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$A$ 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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$640$
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$320$
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$1280$
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$160$

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Motion in a Straight Line

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Uniformly Accelerated Motion

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$A$ particle moves in a straight line with a constant acceleration. It changes its velocity from $10 \, m/s$ to $20 \, m/s$ while passing through a distance of $135 \, m$ in $t$ seconds. The value of $t$ is..........$s$.

MediumAIPMT 2008

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DifficultJEE MAIN 2021

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MediumNEET 2021

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$A$ lift performs the first part of its ascent with uniform acceleration $a$ and the remaining with uniform retardation $2a$. If $t$ is the total time of ascent,find the total height of the shaft.

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$A$ 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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$A$ particle moves in a straight line with a constant acceleration. It changes its velocity from $10 \, m/s$ to $20 \, m/s$ while passing through a distance of $135 \, m$ in $t$ seconds. The value of $t$ is..........$s$.

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