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The distance moved by the particle in time $' t'$ is given by $S = t^{3} - 4 t^{2} + 4 t + 7.$ At the instant when its acceleration is '0' then velocity is

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$\frac{- 4}{3} unit/sec$

$\frac{6}{5} unit/sec$

$48 unit/sec$

$- 48 unit / \sec$

answer is B.

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Detailed Solution

$S = t^{3} - 4 t^{2} + 4 t + 7$

$v = 3 t^{2} - 8 t + 4$

$a = 6 t - 8 = 0$

$t = \frac{8}{6} = \frac{4}{3}$

$V = 3 (\frac{16}{9}) - 8 (\frac{4}{3}) + 4$

$= \frac{16 - 32}{3} + 4 = \frac{16 - 32 + 12}{3} = \frac{- 4}{3}$

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