relative motion

Question

In a river of 20 m width.Half part of river flows with speed of 10m/s and remaining half part flows with 20m/s.as shown in figure.A man starts to swim from A and reaches to point B in 10 sec .Man swims with speed v_m  with respect to river at an angle thita  with line AB. find thita angle --

ans :arctan(15/2)

Answer 1

Here the component of velocity of swimmer in AB direction is always V_mcosθ.

As the swimmer  crosses the river of breadth 20m in 10sec we can write

10* V_mcosθ.=20

=> V_mcosθ.=2........[1]

Again the net displacement along the stream is zero and  each 10m breadth is crossed by the swimmer in 5sec.

So we can write

(V_msinθ.-10)*5 +(V_msinθ.-20)*5=0

=> 2V_msinθ.= 30

=> V_msinθ.= 15.....[2]

Dividing [2] by [1] we get

=> tanθ= 15/2

=> θ.= tan^-1(15/2)