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(D) Given that the distance $s$ is proportional to $t^{1/2}$,we can write $s = k t^{1/2}$,where $k$ is a constant.The velocity $v$ is the first deriva

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Decreasing retardation

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(D) Given that the distance $s$ is proportional to $t^{1/2}$,we can write $s = k t^{1/2}$,where $k$ is a constant.The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(k t^{1/2}) = \frac{1}{2} k t^{-1/2}$.The acceleration $a$ is the second derivative of displacement with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(\frac{1}{2} k t^{-1/2}) = -\frac{1}{4} k t^{-3/2}$.The negative sign indicates that the acceleration is in the opposite direction of motion,which represents retardation (deceleration).As time $t$ increases,the magnitude of acceleration $|a| = \frac{k}{4 t^{3/2}}$ decreases.Therefore,the motion is characterized by decreasing retardation.

The distance travelled by a particle is directly proportional to $t^{1/2}$,where $t =$ time elapsed. What is the nature of motion?

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