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(A) The displacement is given by $s = t^3 - 3t^2 + 2$.The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt

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(A) The displacement is given by $s = t^3 - 3t^2 + 2$.The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = 3t^2 - 6t$.The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d^2s}{dt^2} = 6t - 6$.Set the acceleration to zero to find the time $t$:$6t - 6 = 0 \implies t = 1 \, s$.Now,substitute $t = 1 \, s$ into the displacement equation to find the displacement at that instant:$s = (1)^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0 \, m$.

$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 3t^2 + 2) \, m$. The displacement when the acceleration becomes zero is ........ $m$.

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