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(A) The distance covered by the particle is given by $s = t + 6t^2 - t^3$.The velocity $v$ is the first derivative of distance with respect to time $t

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$2$

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(A) The distance covered by the particle is given by $s = t + 6t^2 - t^3$.The velocity $v$ is the first derivative of distance with respect to time $t$:$v = \frac{ds}{dt} = \frac{d}{dt}(t + 6t^2 - t^3) = 1 + 12t - 3t^2$.The acceleration $a$ is the second derivative of distance with respect to time $t$:$a = \frac{dv}{dt} = \frac{d}{dt}(1 + 12t - 3t^2) = 12 - 6t$.To find the time when the acceleration is zero,set $a = 0$:$12 - 6t = 0$$6t = 12$$t = 2 	ext{ seconds}$.Thus,the acceleration is zero after $2$ seconds.

$A$ particle moves along a straight line such that its distance $s$ in time $t$ seconds is given by $s = t + 6t^2 - t^3$. After what time is the acceleration zero?

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