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.

A swimmer moves across the Cauvery river of 750 m wide. The velocity o

1.	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.
Answer» Solution :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`