Gujarati
Hindi
2.Motion in Straight Line
hard

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

A

$1/x^3$

B

$-t/x^3$

C

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

D

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

Solution

$x^{2}=1+t^{2}$

$\text { or } \quad x=\left(1+t^{2}\right)^{1 / 2}$

${\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{2}\left(1+\mathrm{t}^{2}\right)^{-1 / 2} \cdot 2 \mathrm{t}=\mathrm{t}\left(1+\mathrm{t}^{2}\right)^{-1 / 2}}$

${\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\mathrm{t}\left(-\frac{1}{2}\right)\left(1+\mathrm{t}^{2}\right)^{-3 / 2} \cdot 2 \mathrm{t}+\left(1+\mathrm{t}^{2}\right)^{-1 / 2}}$

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

Standard 11
Physics

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