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(D) Given the position function $S(t) = t^3 - 3t^2 + 4t - 2$. Velocity $v(t)$ is the derivative of $S(t)$ with respect to $t$: $v(t) = \frac{dS}{dt} =

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$1$

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(D) Given the position function $S(t) = t^3 - 3t^2 + 4t - 2$. Velocity $v(t)$ is the derivative of $S(t)$ with respect to $t$: $v(t) = \frac{dS}{dt} = 3t^2 - 6t + 4$. Acceleration $a(t)$ is the derivative of $v(t)$ with respect to $t$: $a(t) = \frac{dv}{dt} = 6t - 6$. Set acceleration to zero to find the time $t$: $6t - 6 = 0 \implies t = 1 	ext{ second}$. Now,substitute $t = 1$ into the velocity function: $v(1) = 3(1)^2 - 6(1) + 4 = 3 - 6 + 4 = 1 	ext{ m/s}$. Thus,the velocity of the particle when its acceleration is zero is $1 	ext{ m/s}$.

$A$ particle is moving along a line according to the law $S = t^3 - 3t^2 + 4t - 2$,where $S$ is measured in meters and $t$ is measured in seconds. Then the velocity (in $m/s$) of the particle when its acceleration is zero is:

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