The position of the particle is given by $x (t) = (4 t^{2} - 3 t + 2 t^{3})$ , its acceleration will be

4 t + 2 t^{2}

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12 t^{2} + 8 t

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8 t - 3 + 6 t^{2}

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12 t + 8

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Solution

The correct option is D $12 t + 8$
$a = \frac{d v}{d t} = \frac{d^{2} x}{d t^{2}} = \frac{d^{2} (4 t^{2} - 3 t + 2 t^{3})}{d t^{2}} = 12 t + 8$

Suggest Corrections

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