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(A) Given the equation of motion: $s = 2t^3 - 9t^2 + 12t$. To find the velocity $v$,we differentiate $s$ with respect to $t$: $v = \frac{ds}{dt} = \fr

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$3/2 	ext{ } sec$

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(A) Given the equation of motion: $s = 2t^3 - 9t^2 + 12t$. To find the velocity $v$,we differentiate $s$ with respect to $t$: $v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 9t^2 + 12t) = 6t^2 - 18t + 12$. To find the acceleration $a$,we differentiate the velocity $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 18t + 12) = 12t - 18$. For the acceleration to be zero,we set $a = 0$: $12t - 18 = 0$. Solving for $t$: $12t = 18$,which gives $t = \frac{18}{12} = \frac{3}{2} 	ext{ } sec$. Thus,the acceleration of the particle will be zero after $3/2 	ext{ } sec$.

The equation of motion of a particle moving along a straight line is $s = 2t^3 - 9t^2 + 12t$,where the units of $s$ and $t$ are $cm$ and $sec$. The acceleration of the particle will be zero after:

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