Question

A man can swim at the rate of `5 km h^-1` in still water. A `1 - km` wide river flows at the rate of `3 km h^-1` The man wishes to swim across the river directly opposite to the starting point.
(a) Along what direction must the man swim ?
(b) What should be his resultant velocity ?
( c) How much time will he take to cross the river ?

Answer 1

(a) Velocity of man with respect to river water, `v = 5 km h^-1`. This is greater than the river flow velocity. Therefore, he can cross the river directly (along the shortest path or no drift condition from flow velocity). The angle of swim,
`theta = (pi)/(2) + sin^-1 ((u)/(v))`
=`90^@ + sin^-1 ((u)/(v))`
=`90^@ + sin^-1 ((3)/(5)) = 90^@ + 37^@ = 127^@` w.r.t. the river flow
or `37^@` w.r.t. perpendicular in upstream direction.
.
(b) Resultant velocity or velocity of mass will be
`v_m = sqrt(v^2 - u^2) = sqrt(5^2 - 3^2) = 4 k=km h^-1`
In the direction perpendicular to the river flow.
( c) time taken to cross the river
`t = (d)/(sqrt(v^2 - u^2)) = (1 km)/(4 km h^-1) =(1)/(4) h = 15 min`.