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

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$-3$

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(C) Given the displacement function $s = t^3 - 3t^2$.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: $a = \frac{dv}{dt} = 6t - 6$.Set the acceleration to zero to find the time $t$: $6t - 6 = 0 \implies t = 1 	ext{ second}$.Now,substitute $t = 1$ into the velocity equation to find the velocity at that instant:$v = 3(1)^2 - 6(1) = 3 - 6 = -3 	ext{ m/s}$.

If the distance $s$ metres traversed by a particle in $t$ seconds is given by $s = t^3 - 3t^2$,then the velocity of the particle when the acceleration is zero,in $m/s$ is:

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