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Question

$x^{2}=1+t^{2}$

$	ext { or } \quad x=\left(1+t^{2}ight)^{1 / 2}$

${\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{2}\left(1+\mathrm{t}^{2}ight)^{

Vedclass · Accepted Answer

$\frac{1}{x} - \frac{{{t^2}}}{{{x^3}}}$

Vedclass · Answer

$x^{2}=1+t^{2}$

$	ext { or } \quad x=\left(1+t^{2}ight)^{1 / 2}$

${\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{2}\left(1+\mathrm{t}^{2}ight)^{-1 / 2} \cdot 2 \mathrm{t}=\mathrm{t}\left(1+\mathrm{t}^{2}ight)^{-1 / 2}}$

${\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\mathrm{t}\left(-\frac{1}{2}ight)\left(1+\mathrm{t}^{2}ight)^{-3 / 2} \cdot 2 \mathrm{t}+\left(1+\mathrm{t}^{2}ight)^{-1 / 2}}$

${=\frac{1}{\mathrm{x}}-\frac{\mathrm{t}^{2}}{\mathrm{x}^{3}}}$

A point moves in a straight line so that its displacement $x\,m$ at time $t\,sec$ is given by $x^2 = 1 + t^2$. Its acceleration in $m/sec^2$ at a time $t\,sec$ is

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