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(D) The velocity is given by $v = 2 t^2 e^{-t}$.Acceleration $a$ is the rate of change of velocity with respect to time,$a = \frac{dv}{dt}$.Using the

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Both $(a)$ and $(b)$

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(D) The velocity is given by $v = 2 t^2 e^{-t}$.Acceleration $a$ is the rate of change of velocity with respect to time,$a = \frac{dv}{dt}$.Using the product rule for differentiation: $a = 2 \left[ t^2 \frac{d}{dt}(e^{-t}) + e^{-t} \frac{d}{dt}(t^2) \right]$.$a = 2 [ t^2 (-e^{-t}) + e^{-t} (2t) ]$.$a = 2 e^{-t} (2t - t^2)$.For acceleration to be zero,set $a = 0$:$2 e^{-t} (2t - t^2) = 0$.Since $2 e^{-t}$ is never zero for any finite $t$,we must have $2t - t^2 = 0$.$t(2 - t) = 0$.This gives $t = 0 \ s$ or $t = 2 \ s$.Thus,the acceleration is zero at $t = 0 \ s$ and $t = 2 \ s$.

The velocity $v$ of a body moving along a straight line varies with time $t$ as $v = 2 t^2 e^{-t}$,where $v$ is in $m/s$ and $t$ is in $s$. The acceleration of the body is zero at $t =$

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