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. The ratio of times of motion of boats τAτB is x10 if the velocity of each boat with respect to water is η=1.2 times greater than the stream velocity. Find x rounded off to the nearest integer.

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Solution

Let l be the distance covered by the boat A along the river as well as by the boat B across the river. Let v0 be the stream velocity and v′ the velocity of each boat with respect water. Therefore time taken by the boat A in its journey

tA=lv′+v0+lv′−v0

and for the boat B tB=l√v′2−v20+l√v′2−v20=2l√v′2−v20

Hence, tAtB=v′√v′2−v20=η√η2−1(whereη=v′v)

On substitution tAtB=1.8

tA=lv′+v0+lv′−v0

and for the boat B tB=l√v′2−v20+l√v′2−v20=2l√v′2−v20

Hence, tAtB=v′√v′2−v20=η√η2−1(whereη=v′v)

On substitution tAtB=1.8

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