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(C) Given the velocity equation: $v(t) = 6t^2 - 6t^3$.To find the extrema,we calculate the first derivative with respect to time: $\frac{dv}{dt} = 12t

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minimum velocity is zero

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(C) Given the velocity equation: $v(t) = 6t^2 - 6t^3$.To find the extrema,we calculate the first derivative with respect to time: $\frac{dv}{dt} = 12t - 18t^2$.Setting $\frac{dv}{dt} = 0$,we get $6t(2 - 3t) = 0$,which gives critical points at $t = 0 \ s$ and $t = 2/3 \ s$.Now,we find the second derivative: $\frac{d^2v}{dt^2} = 12 - 36t$.Evaluating at $t = 0$: $\frac{d^2v}{dt^2} = 12 - 36(0) = 12 > 0$. Since the second derivative is positive,$t = 0 \ s$ corresponds to a local minimum.Evaluating at $t = 2/3$: $\frac{d^2v}{dt^2} = 12 - 36(2/3) = 12 - 24 = -12 At $t = 0 \ s$,the velocity $v = 6(0)^2 - 6(0)^3 = 0 \ m/s$. Thus,the minimum velocity is $0 \ m/s$.

The velocity $v$ of a particle is given by the equation $v = 6t^2 - 6t^3$,where $v$ is in $m/s$ and $t$ is time in $s$. Then:

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