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Question Number 29907 by ajfour last updated on 13/Feb/18

Commented by ajfour last updated on 13/Feb/18

$A b o y s w i m s a t s p e e d v i n s t i l l$ $w a t e r, w a l k s o n g r o u n d w i t h$ $s p e e d w > v . I f h e n e e d s t o c r o s s$ $a r i v e r o f w i d t h b, w a t e r f l o w i n g$ $a t s p e e d u, f i n d t h e m i n i m u m$ $t i m e i n w h i c h h e c a n r e a c h B,$ $s t a r t i n g f r o m A .$
Commented by 33 last updated on 13/Feb/18

$s i r, i s n t t h e r e a n y r e l a t i o n g i v e n$ $b e t w e e n u, v & w o t h e r t h a n$ $t h e i n e q u a l i t y ? l i k e t h e r a t i o o r$ $s o m e t h i n g ?$
Commented by 33 last updated on 13/Feb/18

$i t h i n k i t m i g h t b e r e q u i r e d .$
Commented by ajfour last updated on 13/Feb/18

$A n s w e r i s t o b e o b t a i n e d i n$ $t e r m s o f u, v, w, b .$
Commented by 33 last updated on 13/Feb/18

$o k$
	Answered by mrW2 last updated on 13/Feb/18

	$v_{x} = u - v \sin θ$ $v_{y} = v \cos θ$ $t_{1} = \frac{b}{v_{y}} = \frac{b}{v \cos θ}$ $t_{2} = \frac{v_{x} t_{1}}{w}$ $t = t_{1} + t_{2} = (1 + \frac{v_{x}}{w}) t_{1} = (1 + \frac{u - v \sin θ}{w}) \frac{b}{v \cos θ}$ $t = \frac{b}{w} \times \frac{(\frac{w + u}{v} - \sin θ)}{\cos θ} = \frac{b}{w} (\frac{β - \sin θ}{\cos θ})$ $\frac{d t}{d θ} = \frac{b}{w} [- 1 + \frac{(β - \sin θ) \sin θ}{\cos^{2} θ}] = \frac{β b}{w} \times \frac{\sin θ - \frac{1}{β}}{\cos^{2} θ}$ $s i n c e v < w, \frac{1}{β} = \frac{v}{w + u} < 1$ $\frac{d t}{d θ} = 0 \Rightarrow \sin β - \frac{1}{β} = 0$ $\Rightarrow \sin θ = \frac{1}{β} = \frac{v}{w + u}$ $\Rightarrow θ = \sin^{- 1} (\frac{v}{w + u})$ $\cos θ = \sqrt{1 - {(\frac{v}{w + u})}^{2}}$ $t_{m i n} = \frac{b (\frac{w + u}{v} - \frac{v}{w + u})}{w \sqrt{1 - {(\frac{v}{w + u})}^{2}}} = \frac{b \sqrt{(w + u + v) (w + u - v)}}{w v}$
	Commented by ajfour last updated on 13/Feb/18

	$E x c e l l e n t S i r, a n d e v e n q u i c k!$
	Commented by 33 last updated on 13/Feb/18

	$n i c e o n e s i r$
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