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.

Answer 1

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

$⟹v\sin \theta =u$

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

$⟹\theta ={\sin }^{-1}\frac{3}{4}$

Resultant velocity is along AB, and is $v_{res}=v\cos \theta =20\times \sqrt{1-{\left(\frac{3}{4}\right)}^2}$

$⟹v_{res}=5\sqrt{7}km/h$ along AB.

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

$⟹T=3min38sec$