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 $\mathrm{t}$ is $x^{-\mathrm{n}}$ where $\mathrm{n}=$ ___________

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{x}^2=1+\mathrm{t}^2 \ & 2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t} \ & \mathrm{xv}=\mathrm{t} \ & \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v} \frac{\mathrm{dx}}{\mathrm{dt}}=1 \ & \mathrm{x} \cdot \mathrm{a}+\mathrm{v}^2=1 \ & \mathrm{a}=\frac{1-\mathrm{v}^2}{\mathrm{x}}=\frac{1-\mathrm{t}^2 / \mathrm{x}^2}{\mathrm{x}} \ & \mathrm{a}=\frac{1}{\mathrm{x}^3}=\mathrm{x}^{-3}\end{aligned}$

A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^2=1+t^2$.…

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