A particle moves in a straight line so that its displacement at any time is given by . Its acceleration at any time is where n

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Solution

Q.
A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^{2} = 1 + t^{2}$ . Its acceleration at any time $t$ is $x^{- n}$ where $n =$ _____. [2024]

Ans.
(3) $x^{2} = 1 + t^{2} \Rightarrow x = \sqrt{1 + t^{2}}$

   $v = \frac{d x}{d t} = \frac{1}{2 \sqrt{1 + t^{2}}} \cdot 2 t \Rightarrow v = \frac{t}{\sqrt{1 + t^{2}}}$

   $a = \frac{d v}{d t} = \frac{\sqrt{1 + t^{2}} (1) - t \cdot \frac{1}{2} {(t^{2} + 1)}^{- 1 / 2} \cdot 2 t}{(1 + t^{2})}$

   $a = \frac{\sqrt{1 + t^{2}} - \frac{t^{2}}{\sqrt{1 + t^{2}}}}{(1 + t^{2})} = \frac{t^{2} + 1 - t^{2}}{{(t^{2} + 1)}^{3 / 2}}$

   $\Rightarrow a = \frac{1}{{(1 + t^{2})}^{3 / 2}} = \frac{1}{x^{3}} = x^{- 3} \Rightarrow n = 3$

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