I. E. Irodov Solution 1.7

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: 1.7
Two swimmers leave point \(A\) on one bank of the river to reach point \(B\) lying right across on the other bank. One of them crosses the river along the straight line \(AB\) while the other swims at right angles to the stream and then walks the distance that he has right angles to the stream and then walks the distance that he has been carried away by the stream to get to point \(B\). What was the velocity \(u\) of the walking if both swimmers reached the destination simultaneously? The stream velocity \({v_0} = 2.0km/h\) and the velocity \(v'\) of each swimmer with respect to water equals \(2.5km/h\).

Solution: 1.7

Let us call the swimmer who swam directly from \(A\) to \(B\) as \({S_1}\) and the other swimmer as \({S_2}\). \({S_1}\), in-order to swim directly to \(B\) from \(A\), had to oppose the flow of the river and completely annul it. This means that \({S_1}\) must have had to swim at an angle \(\theta \) from the line \(AB\) such that, the velocity component along the direction of river flow must be exactly equal and opposite to that of the river flow.
\(v'\sin \theta = {v_0}\)
or, \(\sin \theta = \frac{{v'}}{{{v_0}}}\) ... ... ... ... (i)

Thus we have, the component of velocity along the direction \(AB\) of the swimmer \({S_1}\) is given by
\(v'\cos \theta = v'\sqrt {1 - \frac{{v_0^2}}{{v{'^2}}}} \) ... ... ... ... (ii)
If the length \(AB\) is \(d\) then the time taken by swimmer \({S_1}\) to reach \(B\) is given by,
\({t_{{S_1}}} = \frac{d}{{v'\sqrt {1 - \frac{{v_0^2}}{{v{'^2}}}} }}\) ... ... ... ... (iii)

Swimmer \({S_2}\) however, swims perpendicular to the river's flow and hence will be swept off to the right by the river. His speed in the direction perpendicular to the river will be \(v'\) and the river will push him to the right with a speed \({v_0}\). Thus, swimmer \({S_2}\) will take \(\frac{d}{{v'}}\) time to reach the other shore at point \(C\) as seen in the figure. Since, he has been drifting to the right with a speed \(v'\) to the right, \(BC\) will be equal to \(\frac{{d{v_0}}}{{v'}}\). He will thus take \(\frac{{d{v_0}}}{{v'u}}\) time to reach \(B\) from \(C\). This means that the total time swimmer \({S_2}\) takes to reach \(B\) via \(C\) will be equal to,
\({t_{{S_2}}} = \frac{d}{{v'}} + \frac{{d{v_0}}}{{v'u}}\) ... ... ... ... (v)

Since both the swimmers reach at the same time, we have,
\(\frac{d}{{v'\sqrt {1 - \frac{{v_0^2}}{{v{'^2}}}} }} = \frac{d}{{v'}} + \frac{{d{v_0}}}{{v'u}}\)
or, \(u = \frac{{{v_0}}}{{\left[ {\frac{1}{{\sqrt {\left( {1 - \frac{{v_0^2}}{{v{'^2}}}} \right)} }} - 1} \right]}}\)
Here,
\({v_0} = 2km/h\) and \(v' = 2.5km/h\)
Thus, \(u = 3km/h\)