Acceleration of a particle which is at rest at x = 0 is a^→ = (4 − 2x) i^∧ . Select the correct alternative(s). (a) Maximum speed of the particle is 4 units (b) Particle further comes to rest at x = 4 (c) Particle oscillates about x = 2 (d) Particle will continuously accelerate along the x-axis.


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Question Number 20842 by Tinkutara last updated on 04/Sep/17

$Acceleration of a particle which is at$ $rest at x = 0 is \vec{a} = (4 - 2 x) \overset{\land}{i} . Select$ $the correct alternative (s) .$ $(a) Maximum speed of the particle is$ $4 units$ $(b) Particle further comes to rest at$ $x = 4$ $(c) Particle oscillates about x = 2$ $(d) Particle will continuously accelerate$ $along the x - axis .$
	Answered by ajfour last updated on 09/Sep/17

	$\frac{v_{x} {d v}_{x}}{d x} = 4 - 2 x$ $\Rightarrow \int_{0}^{v_{x}} v_{x} {d v}_{x} = \int_{0}^{x} (4 - 2 x) d x$ $\Rightarrow \frac{v_{x}^{2}}{2} = 4 x - x^{2}$ $\Rightarrow v_{x}^{2} = 2 [4 - {(x - 2)}^{2}]$ $\Rightarrow p a r t i c l e c o m e s t o r e s t a t x = 4 .$ $m a x i m u m s p e e d a t x = 2 a n d i s$ ${(v_{x})}_{m a x} = 2 \sqrt{2} u n i t s .$ $\Rightarrow \int_{0}^{x} \frac{d x}{\sqrt{2^{2} - {(x - 2)}^{2}}} = \pm \sqrt{2} \int_{0}^{t} d t$ $\Rightarrow \sin^{- 1} (\frac{x - 2}{2}) - \sin^{- 1} (- 1) = \pm \sqrt{2} t$ $\Rightarrow \sin^{- 1} (\frac{x - 2}{2}) = \pm \sqrt{2} t - π / 2$ $x = 2 + 2 \sin (\pm \sqrt{2} t - π / 2)$ $o r x = 2 - \cos (\sqrt{2} t)$ $i s t h e e q u a t i o n o f m o t i o n .$ $\Rightarrow p a r t i c l e o s c i l l a t e s a b o u t x = 2 .$ $(b) a n d (c) a r e t h e c o r r e c t o p t i o n s .$
	Commented by ajfour last updated on 09/Sep/17

	$t h a n k s, i s h a l l b e c a r e f u l .$
	Commented by alex041103 last updated on 09/Sep/17

	$Y o u h a v e a m i s t a k e$ $\Rightarrow \int_{0}^{x} \frac{d x}{\sqrt{2^{2} - {(x - 2)}^{2}}} = \pm \sqrt{2} \int_{0}^{t} d t$ $\Rightarrow x (t) = 2 - 2 c o s (\sqrt{2} t)$
	Commented by Tinkutara last updated on 04/Sep/17

	$Thank you very much Sir!$
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