Answer to Question #137405 in Mechanics | Relativity for Viccy

Question #137405
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 = 1.2 times greater than the stream velocity.
1
Expert's answer
2020-10-12T07:49:58-0400

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}=1.2"

Therefore ratio becomes

"\\frac{t_A}{t_B}=\\frac{1.2}{\\sqrt{(1.2)^2-1}}=1.8"

Therefore ratio of time taken by both boats is 1.8


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