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Problem Solving

A river 800 m...

Question

A river 800 m wide close at a rate of 15km/hr. A swimmer who can swim at 20 km/h in still water wishes to cross the given straight.
(i) Along what direction must he strike
(ii) What should be his resultant velocity
(iii) How much time he would take.

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Solution

There is no displacement in $X -$ direction, so net velocity in this direction should be $0$ .

$⟹ v sin θ = u$

$⟹ sin θ = \frac{15}{20} = \frac{3}{4}$

$⟹ θ = {sin}^{- 1} \frac{3}{4}$

Resultant velocity is along AB, and is $v_{r e s} = v cos θ = 20 \times \sqrt{1 - {(\frac{3}{4})}^{2}}$

$⟹ v_{r e s} = 5 \sqrt{7} k m / h$ along AB.

Time, $T = \frac{d}{v_{r e s}} = \frac{800}{5 \sqrt{7} \times \frac{5}{18}} = 217.7 s$

$⟹ T = 3 m i n 38 s e c$

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