A boy can swim with a speed of 26m/s in still water.He wants to swim across a 150m river from a point A to point B which is directly opposite the other side of the river.The river flows with a speed of 10m/s. i)if he always swim in the direction parallel to AB,find how far he lands downstream of B. ii)In what direction relative to the bank must he swim so as to cross directly from A to B.


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Question Number 30738 by NECx last updated on 25/Feb/18

$A b o y c a n s w i m w i t h a s p e e d o f$ $26 m / s i n s t i l l w a t e r . H e w a n t s t o$ $s w i m a c r o s s a 150 m r i v e r f r o m$ $a p o i n t A t o p o i n t B w h i c h i s$ $d i r e c t l y o p p o s i t e t h e o t h e r s i d e$ $o f t h e r i v e r . T h e r i v e r f l o w s w i t h$ $a s p e e d o f 10 m / s .$ $i) i f h e a l w a y s s w i m i n t h e$ $d i r e c t i o n p a r a l l e l t o A B, f i n d h o w$ $f a r h e l a n d s d o w n s t r e a m o f B .$ $i i) I n w h a t d i r e c t i o n r e l a t i v e t o$ $t h e b a n k m u s t h e s w i m s o a s t o$ $c r o s s d i r e c t l y f r o m A t o B .$
Commented by NECx last updated on 25/Feb/18

$p l e a s e h e l p$
	Answered by mrW2 last updated on 26/Feb/18

	$a s s u m e t h e b o y i s a s u p e r m a n w h o$ $c a n r e a l l y s w i m w i t h a s p e e d o f$ $26 m / s (= 93.6 k m / h) .$ $(i)$ $a = \frac{150}{26} \times 10 = 57.7 m$ $(i i)$ $\cos θ = \frac{10}{26}$ $\Rightarrow θ = \cos^{- 1} \frac{10}{26} = 67.4 °$
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