Two boats A and B move away from a bauy anchored at the middle of a river along the mutually perpendicular straight lines,the boat A along the river and the boat B across the river.Having moved off equal distance from the bauy the boats returned.Find the ratio of times of boats Ta/Tb if the velocity of both the boats with respect to water is n=1.2 times greater than the stream velocity.

Considering for the boat B to have it's velocity in such a direction so that it travels straight across the river on the perpendicular bisector of river's axis(path of boat A).Then for the same distance travelled D

Value of T_a=(D/V)*[1/2.2+1/.2]       For Downstream velocity of boat A=vel. of A+vel. of stream ;vel. of stream=V

                                                            while Upstream vel. of A=vel. of A-vel. of stream ;vel. of A (&B)=1.2*vel. of stream

T_b=(D/V)*[2/0.663]                Magnitude of vel. of boat B is 1.2V of whose component in the direction of flow of river

                                                                                                                            =stream vel.=V

                                              or,(1.2V)Cosθ=V       (θ=angle made by 1.2V with flow of river but in such a way that it is

                                                                                              opposite to stream vel. V)

                                                  From above we get (1.2V)Sinθ which is the across component to cross distance D.

  T_a/T_b=[1/2.2+1/.2]/[2/0.663] =1.81