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Acceleration

The position ...

Question

The position of the particle is given by $x = 2 t^{3} - 4 t^{2} + 5$ in $m$ . The acceleration of the particle at $5 sec$ is

45 m / s^{2}

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48 m / s^{2}

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40 m / s^{2}

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52 m / s^{2}

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Solution

The correct option is D $52 m / s^{2}$
As we know that the acceleration of the particle is given by
$a = \frac{d v}{d t}$ and also, $v = \frac{d x}{d t}$
Given, $x = 2 t^{3} - 4 t^{2} + 5$
So, $v = \frac{d x}{d t} = \frac{d (2 t^{3} - 4 t^{2} + 5)}{d t}$
$\Rightarrow v = 6 t^{2} - 8 t$
also, $a = \frac{d v}{d t} = \frac{d (6 t^{2} - 8 t)}{d t} = 12 t - 8$
Thus, the acceleration of the particle is $12 \times 5 - 8 = 52 m / s^{2}$

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