A man can swim with a speed of 4 $k m h^{- 1}$ in still water. He crosses a river 1 km wide that flows steadily at 3 $k m h^{- 1}$ . If he makes his strokes normal to the river current, how far down the river does he go when he reaches the other bank?

500 m

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600 m

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750 m

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850 m

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Solution

The correct option is C 750 m
Time to cross the river, $t = \frac{W i d t h o f r i v e r}{S p e e d o f m a n} + \frac{1 k m}{4 k m h^{- 1}}$
$= \frac{1}{4} h$
Distance moved along the river in time t,
$s = v_{r} \times t = 3 k m h^{- 1} \times \frac{1}{4} h$
$= \frac{3}{4} k m = 750 m$

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A man can swim with a speed of 4 kmh−1 in still water. He crosses a river 1 km wide that flows steadily at 3 kmh−1. If he makes his strokes normal to the river current, how far down the river does he go when he reaches the other bank?

The correct option is C 750 mTime to cross the river, t=Width of riverSpeed of man+1km4kmh−1 =14hDistance moved along the river in time t, s=vr×t=3kmh−1×14h=34km=750m

A man can swim with a speed of 4 $k m h^{- 1}$ in still water. He crosses a river 1 km wide that flows steadily at 3 $k m h^{- 1}$ . If he makes his strokes normal to the river current, how far down the river does he go when he reaches the other bank?

The correct option is C 750 m
Time to cross the river, $t = \frac{W i d t h o f r i v e r}{S p e e d o f m a n} + \frac{1 k m}{4 k m h^{- 1}}$
$= \frac{1}{4} h$
Distance moved along the river in time t,
$s = v_{r} \times t = 3 k m h^{- 1} \times \frac{1}{4} h$
$= \frac{3}{4} k m = 750 m$