A body starts from the origin and moves along the $x - a x i s$ such that velocity at any instant is given by $(4 t^{3} - 2 t)$ where $t$ is in second and velocity is in $m / s$ . What is the acceleration of the particle, when it is $2 m$ from the origin?

28 m / s^{2}

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22 m / s^{2}

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12 m / s^{2}

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10 m / s^{2}

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Solution

The correct option is B $22 m / s^{2}$
Given that $v = (4 t^{3} - 2 t)$
$x = \int v d t, x = t^{4} - t^{2} + C$ at $t = 0, x = 0$
$\Rightarrow C = 0$
When particle is $2 m$ away from the origin, then
$2 = t^{4} - t^{2} \Rightarrow t^{4} - t^{2} - 2 = 0$
$\Rightarrow t = \sqrt{2} s$
$a = \frac{d v}{d t} = \frac{d}{d t} (4 t^{3} - 2 t)$
$a = 12 t^{2} - 2$
For $t = \sqrt{2} s$
$a = 12 \times {(\sqrt{2})}^{2} - 2 = 22 m / s^{2}$

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