Answer to Question #137747 in Field Theory for viccy

Question

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 τA/τB if the velocity of each boat with respect to water is η = 1.4 times greater than the stream velocity.

Answer 1

Solution

Let boat A covered $l$ distance, $v$ is stream velocity, $v '$ is each boat velocity with respect to water.

So time is taken by Boat A is given by

$t_A=\frac{l}{v'+v}+\frac{l}{v'-v}$

Similarly for Boat B is

$t_B=\frac{l}{\sqrt{v'^2-v^2}}+\frac{l}{\sqrt{v'^2-v^2}}=\frac{2l}{\sqrt{v'^2-v^2}}$

Therefore

Ratio of time taken by Boats is given by

$\frac{t_A}{t_B}=\frac{v'}{\sqrt{v'^2-v^2}}=\frac{\eta}{\sqrt{\eta^2-1}}$

Where $\eta=\frac{v'}{v}=1.4$

Therefore ratio becomes

$\frac{t_A}{t_B}=\frac{1.4}{\sqrt{(1.4)^2-1}}=1.42$

Therefore ratio of time taken by both boats is 1.42