Two boats, A and B, move away from a buoy 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 an equal distance from the buoy the boats returned. Find the ratio of times of motion of boats `tau_A//tau_B` if the velocity of each boat with respect to water is `eta=1.2` times greater than the stream velocity.

Options

A. 2.3
B. 1.8
C. 0.5
D. 0.2

Answer

B

Explanation

Suppose the stream velocity is `v_s=v`, then the velocity of each boat with respect to water is `v_b=1.2` v. Let each boat travel a distance l. Then for boat A, time of motion

`tau_A=l/(v_b+v_s)+l/(v_b-v_s)`
`=[l/(1.2v+v)+l/(1.2v-v)]=(60l)/(11v)`...(i)
For the boat B, time of motion
`tau_B=l/sqrt(v_b^2-v_s^2)+l/sqrt(v_b^2-v_s^2)=(2l)/sqrt(v_b^2-v_s^2)`...(ii)
`=(2l)/sqrt((1.2v)^2-v^2) =(3.01l)/v`
The ratio `tau_A/tau_B=(60l//11v)/(3.01l//v) approx` 1.8