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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 ta/tb if the velocity of each boat with respect to water is η = 1.2 times greater than the stream velocity.Answer in terms of η

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 ta/tb if the velocity of each boat with respect to water is η = 1.2 times greater than the stream velocity.Answer in terms of η

Grade:12

1 Answers

Swapnil Saxena
102 Points
9 years ago

The answer to this question is n/((n)2-1)1/2 (n stands for η)

Solution:

Assume the distance they transveresed=d

For A the velocity when he  is gong with  the flow of the river is nv+v and when he is coming against it is nv-v

so the time required is (d/nv+v)+(d/nv-v) = 2nd/((nv)2-v2)

For B, the velocity in going and getting back is ((nv)2-v2)1/2

Fo the time is 2d/((nv)2-v2)1/2

The TA/Tb= (2nd/(nv2-v2))/(2d/((nv)2-v2)1/2)=n/((nv)2-v2)1/2=n/((n)2-1)1/2

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