The displacement $x$ of a particle along a straight line at time $t$ is given by $x = a_{0} + a_{1} t + a_{2} t^{3}$ . The acceleration $a$ of the particle at time $t = 5$ is

15 a_{2}

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30 a_{2}

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a_{1} + 30 a_{2}

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6 a_{2}

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Solution

The correct option is B $30 a_{2}$
Instantaneous acceleration is given by, $a = \frac{d v}{d t}$ and $v = \frac{d x}{d t} \Rightarrow v = \frac{d (a_{0} + a_{1} t + a_{2} t^{3})}{d t} = a_{1} + 3 a_{2} t^{2}$
$∴ a = \frac{d (a_{1} + 3 a_{2} t^{2})}{d t} \Rightarrow a = 6 a_{2} t$
Hence, acceleration $a$ at $t = 5 \Rightarrow a = 6 a_{2} \times 5 = 30 a_{2}$

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