Question-29907

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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