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A particle moves according to the law $S = t^{3} - 3 t^{2} + 3 t + 12$ . The velocity when the acceleration is zero is

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$\frac{8}{3}$ unit/sec

10 unit

–3 unit/sec

answer is B.

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Detailed Solution

$S = t^{3} - 3 t^{2} + 3 t + 12 V = S^{1} = 3 t^{2} - 6 t + 3 a = s^{11} = 6 t - 6 = 0 \Rightarrow t = 1 v = 3 {(1)}^{2} - 6 (1) + 3 = 6 - 6 = 0$

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