Let's go boating!

Classical Mechanics Level 4

Two boats A and B move away from a buoy anchor at the middle of the river along the mutually perpendicular straight line ; the boat A along the river and the boat B across the river. Having moved an equal distance from the buoy , the boats returned. Find the ratio of time of motion of the two boats $\frac{t_A}{t_B}$ ; if the velocity of each boat with respect to water is 1.2 times greater than the stream velocity.

3.06 1 2.16 3.27

1 solution

Hatim Zaghloul
May 19, 2014

Assuming the stream speed is V and that the distance they traveled (one way) is D. Tb = 2D/(1.2V). Ta= D/(V+1.2V) + D/(1.2V-V)=D/2.2V +D/0.2V. Ta/Tb = 3.27.

In case of boat b, in order to move in direction perpendicular to boat a it has to move against the current at some angle so that the the resultant of stream velocity and velocity of boat is perpendicular to line of motion of boat a.So by using Pythagoras theorem,resultant velocity of boat b is sq.root of (1.2V)^2-V^2. By the way this is problem of famous book "IE IRODOV Problems in General Physics" and the answer of this question is 1.8.See this (http://irodovsolutionsmechanics.blogspot.in/2007/07/problem-18.html)

satvik pandey - 6 years, 11 months ago

I agree that the answer for the ratio is 1.8 - if both boats have identical thrust and speed of 1.2 times stream velocity. The only way you can arrive at the ratio of the times to be 3.27 is if boat b has enough thrust to achieve a speed of 1.562 times the stream velocity. This would allow boat b to travel at 1.2 times stream velocity perpendicular to the stream current while fighting it. After not seeing the correct answer available, I reluctantly did it the way I assumed the problem creator might have - and got the points. He just analysed the vector velocities incorrectly.

Bob Kadylo - 5 years, 7 months ago

Let's go boating!

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