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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2.Motion in Straight Line

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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 $\mathrm{x}^{-\mathrm{n}}$ where $\mathrm{n}=$ . . . . .

A

$5$

B

$2$

C

$3$

D

$1$

(JEE MAIN-2024)

Solution

$x^2=1+t^2$

$2 x \frac{d x}{d t}=2 t$

$x v=t$

$x \frac{d v}{d t}+v \frac{d x}{d t}=1$

$x \cdot a+v^2=1$

$a=\frac{1-v^2}{x}=\frac{1-t^2 / x^2}{x}$

$a=\frac{1}{x^3}=x^{-3}$

Standard 11

Physics

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