A swimmer moves across the Cauvery river of 750 m wide. The velocity of the swimmer relative to water `(vecv_(SW))` is `1.5ms^(-1)` and directed perpendicular to the water current. The velocity of water relative to the bank `(vecv_(wb))` is `1ms^(-1)`. Calculate the
velocity of the swimmer with respect to the bank of the river `(vecv_(sb))`.
(b) time taken by the swimmer to cross the Cauvery river.

Explanation

We can draw the following picture from the given data in the problem.

The velocity of the swimmer relative to the bank `vecv_(sb)=vecv_(sw)+vecv_(wb)`
Since the swimmer travels in the perpendicular direction against the water current The magnitude is given by
`vecv_(sb)=sqrt(v_(sw)^(2)+v_(wb)^(2))`
`sqrt(1.5^(2)+1^(2))=sqrt(3.25)ms^(-1)~=1.802ms^(-1)`
The direction of the swimmer relative to the bank is given by
`tantheta=(v_(sw))/(v_(wd))=(1.5)/(1)=1.5`
`theta=tan^(-1)(1.5)~~56^(@)`.
(b) The time taken by the swimmer to cross the river is equal to the total distance by the swimmer with velocity `1.802ms^(-1)`
The total distance covered by him,
`d=("width of the river")/(sin56^(@))=(750)/(0.829)=904.7m`
The time taken by the swimmer,
`T=(d)/(v_(ab))=(904.7)/(1.802)=502s`