I. E. Irodov Solution 1.8

Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.

Problem: 1.6
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 ration of times of motion of boats \(\frac{{{t_A}}}{{{t_B}}}\) if the velocity of each boat with respect to water is \(\eta = 1.2\) times greater than the stream velocity.

Solution: 1.6
Let us suppose that the boats \(A\) and \(B\) each traveled a distance \(d\) from the buoy before turning back. Let the boats' speed be \(v\) and the river's speed be \(u\).
For \(A\) while going downstream the river aided the boat, thus it moved with a speed \(\left( {v + u} \right)\). On its upstream journey back to the buoy, however the river was opposing the boat and so its speed was \(\left( {v - u} \right)\). The total time taken by \(A\) thus is given by, \({t_A} = \frac{d}{{\left( {v + u} \right)}} + \frac{d}{{\left( {v - u} \right)}}\) ... ... ... ... (i)

Similar to case for \(B\) to travel perpendicular to the buoy, he must move at an angle to the perpendicular to the buoy, he must move at an angle to the perpendicular. His speed perpendicular to the buoy is thus given by, \(v\sqrt {1 - \frac{{{u^2}}}{{{v^2}}}} \).

This will be the same \(B'\) return journey as well. Thus, total time taken by him is given by,
\({t_B} = \frac{{2d}}{{v\sqrt {1 - \frac{{{u^2}}}{{{v^2}}}} }} = \frac{{2d}}{{\sqrt {{v^2} - {u^2}} }}\) ... ... ... ... (ii)

Thus, from (i) and (ii) we get,
\(\frac{{{t_A}}}{{{t_B}}} = \frac{v}{{\sqrt {{v^2} - {u^2}} }} = \frac{\eta }{{\sqrt {{\eta ^2} - 1} }}\)