I. E. Irodov Solution 1.10

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: 1.10
A boat moves relative to water with a velocity which is \(n = 2.0\) times less than the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting?

Solution: 1.10

The figure shows the components of the velocities of the two bodies along the horizontal and vertical components. The position of the vertically thrown body after time \(t\) will be \(0\hat i + \left( {{v_0}t - \frac{1}{2}g{t^2}} \right)\hat j\), where \(\hat i\) and \(\hat j\) are the unit vectors along the horizontal and vertical directions respectively.

The position of the other body at a function of time will be \({v_o}t\hat i + \left( {{v_0}t\sin \theta - \frac{1}{2}g{t^2}} \right)\hat j\).
Now the distance between the two is simply the euclidean distance given by
\(\sqrt {v_0^2{t^2}{{\cos }^2}\theta + {{\left( {{v_0}t - {v_o}t\sin \theta } \right)}^2}} \)
\( = {v_0}t\sqrt {2\left( {1 - \sin \theta } \right)} \)
\( = 22m\)