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Question Number 142882 by mohammad17 last updated on 06/Jun/21

Commented by mr W last updated on 06/Jun/21

$h o w d e e p i s t h e l a k e ?$ $w h a t a r e t h e u n i t e s o f a a n d v i n$ $a = 10 - 0.01 v^{2}$
	Answered by Olaf_Thorendsen last updated on 07/Jun/21

	$a = 10 - 0.01 v^{2}$ $\frac{a}{10 - 0.01 v^{2}} = 1$ $\frac{d v}{1000 - v^{2}} = \frac{1}{100} d t$ $\frac{1}{10 \sqrt{10}} argtanh (\frac{v}{10 \sqrt{10}}) - \frac{1}{10 \sqrt{10}} \arcsin (\frac{v_{0}}{10 \sqrt{10}}) = \frac{t}{100}$ $v = 10 \sqrt{10} \tanh (\frac{t}{10 \sqrt{10}} + argtanh (\frac{v_{0}}{10 \sqrt{10}}))$ $v = 10 \sqrt{10} \tanh (\frac{t}{10 \sqrt{10}} + 0.516)$ $v_{m a x} = 10 \sqrt{10} = 31.623$
	Answered by mr W last updated on 07/Jun/21

	$a = \frac{d v}{d t} = v \frac{d v}{d y} = 10 - 0.01 v^{2}$ $\frac{v d v}{10 - 0.01 v^{2}} = d y$ $\int_{v_{0}}^{v} \frac{v d v}{10 - 0.01 v^{2}} = \int_{0}^{y} d y$ $\int_{v_{0}}^{v} \frac{d (10 - 0.01 v^{2})}{10 - 0.01 v^{2}} = - 0.02 \int_{0}^{y} d y$ $\ln \frac{10 - 0.01 v^{2}}{10 - 0.01 v_{0}^{2}} = - 0.02 y$ $\frac{10 - 0.01 v^{2}}{10 - 0.01 v_{0}^{2}} = e^{- 0.02 y}$ $v^{2} = \frac{1}{0.01} [10 - (10 - 0.01 v_{0}^{2}) e^{- 0.02 y}]$ $v = 10 \sqrt{10 - (10 - 0.01 v_{0}^{2}) e^{- 0.02 y}}$
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