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

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-42

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-48

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Solution

The correct option is D -42
$x = t^{3} - 12 t^{2} + 6 t + 8$ .
$v = \frac{d x}{d t} = 3 t^{2} - 24 t + 6$
$a = \frac{d v}{d t} = 6 t - 24$

$\Rightarrow 6 t - 24 = 0$

$\Rightarrow t = 4$

So, $v (t = 4) = 3 (4)^{2} - 24 (4) + 6 = 3 (16) - 96 + 6$

$v = - 42$

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