The distance moved by the particular in time $t$ is given by $s = t^{3} - 12 t^{2} + 6 t + 8$ . At the instant, when its acceleration is zero the velocity is

42

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- 42

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48

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- 48

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Solution

The correct option is B $- 42$
Given, $s = t^{3} - 12 t^{2} + 6 t + 8$
$\frac{d s}{d t} = 3 t^{2} - 24 t + 6$

$\frac{d^{2} s}{d t^{2}} = 6 t - 24$
$A c c e l e r a t i o n = 0$
$⟹ 6 t - 24 = 0$
$∴ t = 4$
Required velocity $= \frac{d s}{d t}]_{t = 4}$
$= 3 \times (4)^{2} - 24 \times 4 + 6$
$= 48 - 96 + 6 = - 42 u n i t s$

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