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Q.
A particle moves a distance x in time t according to equation $x^{2}=1+t^{2}$ The acceleration of the particle is
Motion in a Straight Line
Solution:
Given, $x^{2}=t^{2}+1 \dots$(i)
Differentiating with respect to $t$ on both sides, we get
$2 x \frac{d x}{d t}=2 t$ or $x v=t \dots$ (ii)
Again differentiating w.r.t. $t$ on both sides, we get
$x \frac{d v}{d t}+v \frac{d x}{d t}=1 $ or
$ x \frac{d v}{d t}=1-v^{2}$
or $ \frac{d v}{d t}=\frac{1-v^{2}}{x}=\frac{1-\frac{t^{2}}{x^{2}}}{x}$
$=\frac{x^{2}-t^{2}}{x^{3}}=\frac{1}{x^{3}} $ (Using (i, ii))
$\therefore $ Acceleration, $a=\frac{d v}{d t}=\frac{1}{x^{3}}$