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Question Number 60477 by Kunal12588 last updated on 21/May/19

	Answered by tanmay last updated on 21/May/19

	$v \frac{d v}{d x} = - k (L - x)$ $i n t r e g a t i n g$ $\frac{v^{2}}{2} = - k (L x - \frac{x^{2}}{2}) + c$ $v_{} = v_{0} w h e n x = 0$ $\frac{v_{0}^{2}}{2} = c$ $\frac{v^{2}}{2} = - k (L x - \frac{x^{2}}{2}) + \frac{v_{0}^{2}}{2}$ $\frac{{(\frac{v_{0}}{2})}^{2}}{2} = - k (L \times L - \frac{L^{2}}{2}) + \frac{v_{0}^{2}}{2}$ $k (\frac{L^{2}}{2}) = \frac{v_{0}^{2}}{2} - \frac{v_{0}^{2}}{8}$ $k = \frac{2}{L^{2}} \times \frac{3 v_{0}^{2}}{8} = \frac{3 v_{0}^{2}}{4 L^{2}}$ $r e t a r d a t i o n = - k (L - x)$ $i n i t i a l r e t a r d a t i o n = - (\frac{3 v_{0}^{2}}{4 L^{2}}) \times (L - 0)$ $= \frac{- 3 v_{0}^{2}}{4 L}$ $= \frac{- 3 \times 25}{4 \times 1} = \frac{- 75}{4}$ $p l s c h e c k$
	Commented by Kunal12588 last updated on 21/May/19

	$A b s o l u t e l y c o r r e c t s i r p l s c h e c k m y s o l u t i o n$ $s i r c a n w e t a k e x f r o m r i g h t s i d e i m e a n$ $a = - k x$
	Commented by tanmay last updated on 21/May/19

	$y e s . . . .$ $c o m p l i c a t i o n w i l l a r i s e . .$ $i) t i m e v a r i a t i o n f r o m l e f t b u t d i s t s n c e f r o m$ $r i g h t . . .$
	Commented by Kunal12588 last updated on 21/May/19

	$y e s s i r w h e n i d i d t h a t i w a s g e t t i n g a > 0 .$
	Answered by Kunal12588 last updated on 21/May/19

	Commented by tanmay last updated on 21/May/19

	$y e s c o r r e c t$
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