A particle moves in a straight line along x-axis. At t = 0, it is released from rest at x = a. Acceleration of the particle varies as (d^2 x/dt^2 ) = −(k/x^2 ) , where k is a positive constant. Time required by particle to reach the origin will be


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Question Number 22304 by Tinkutara last updated on 15/Oct/17

$A particle moves in a straight line along$ $x - axis . At t = 0, it is released from$ $rest at x = a . Acceleration of the$ $particle varies as \frac{d^{2} x}{{d t}^{2}} = - \frac{k}{x^{2}}, where k$ $is a positive constant .$ $Time required by particle to reach the$ $origin will be$
	Answered by ajfour last updated on 16/Oct/17

	$\frac{d^{2} x}{{d t}^{2}} = \frac{v d v}{d x} = - \frac{k}{x^{2}}$ $\Rightarrow \int_{0}^{v} v d v = - \int_{a}^{x} \frac{k d x}{x^{2}}$ $\Rightarrow \frac{v^{2}}{2} = + k (\frac{1}{x} - \frac{1}{a})$ $o r v = - \sqrt{2 k} \sqrt{\frac{1}{x} - \frac{1}{a}}$ $\Rightarrow \int_{a}^{x} \frac{d x}{\sqrt{\frac{1}{x} - \frac{1}{a}}} = - \sqrt{2 k} \int_{0}^{t} d t$ $t i m e t a k e n t o r e a c h x = 0 b e T$ $\Rightarrow T = \frac{1}{\sqrt{2 k}} \int_{0}^{a} \sqrt{\frac{a x}{a - x}} d x$ $l e t a - x = y^{2} \Rightarrow d x = - 2 y d y$ $x = a - y^{2}; x = 0 \Rightarrow y = \sqrt{a}$ $a n d x = a \Rightarrow y = 0; t h e n$ $T = \frac{1}{\sqrt{2 k}} \int_{\sqrt{a}}^{0} \frac{\sqrt{a (a - y^{2})}}{y} (- 2 y d y)$ $= \sqrt{\frac{2 a}{k}} \int_{0}^{\sqrt{a}} \sqrt{{(\sqrt{a})}^{2} - y^{2}} d y$ $= \sqrt{\frac{2 a}{k}} [\frac{y}{2} \sqrt{a - y^{2}} ∣_{0}^{\sqrt{a}} + \frac{a}{2} \sin^{- 1} (\frac{y}{\sqrt{a}}) ∣_{0}^{\sqrt{a}}]$ $T = \sqrt{\frac{2 a}{k}} (\frac{a}{2} \times \frac{π}{2}) = \frac{π a}{2} \sqrt{\frac{a}{2 k}} .$
	Commented by squidward last updated on 16/Oct/17

	$4 t h l i n e i s i n c o r r e c t t o o$
	Commented by Tinkutara last updated on 15/Oct/17

	$Thank you very much Sir!$
	Commented by squidward last updated on 16/Oct/17

	$y o u r 3 r d l i n e i s w r o n g a n d h o w t h e h e l l$ $d i d y o u g o f r o m 3 r d l i n e t o 4 t h ?$
	Commented by ajfour last updated on 16/Oct/17

	$i s i t f i n e n o w ?$
	Commented by ajfour last updated on 16/Oct/17

	$N o . a f t e r b e i n g r e l e a s e d a t x = a$ $p a r t i c l e d e v e l o p s n e g a t i v e v e l o c i t y .$ $v = \pm \sqrt{2 k} \sqrt{\frac{1}{x} - \frac{1}{a}}$ $s o w e c h o o s e t h e - v e o n e .$
	Commented by Tinkutara last updated on 25/Nov/17

	$W h y w e c h o s e - v e o n e ?$
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