I. E. Irodov Solution 1.1

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.

Question: 1.1
A motorboat going downstream overcame a raft at a point \(A\). \(\tau = 60\min \) later it turned back and after some time passed the raft at a distance \(l = 6.0km\) from the point \(A\). Find the flow velocity assuming the duty of the engine to be constant.



Solution:
Since the raft is floating on the river, it moves at the same velocity as the river. Let the velocity of the river be \({V_r}\). Let the velocity of the boat in still water be \(V\). This means that the boat's downstream velocity as seen from the ground will be \(\left( {V + {V_r}} \right)\) and it upstream velocity as seen from the ground will be \(\left( {V - {V_r}} \right)\).

As seen by an observer on the raft, when the motorboat moves upstream, its velocity will be \(\left( {V + {V_r} - {V_r}} \right) = V\) and when its moving downstream, it will be \(\left( { - V + {V_r} - {V_r}} \right) = - V\). Here, positive sign indicates that, the motorboat is moving away from the raft and negative sign indicates that, the motorboat is moving towards the raft.

This figure shows the Time-Displacement diagram of the motorboat and the raft as seen from the point of view of the raft. The slopes of the both upstream and downstream paths of the motorboat relative to the raft are \(V\) and \( - V\). Let \({t_1}\) be the time takes by the raft to meet the boat after it turns around. As seen from the figure, it is obvious that,
\({t_1} = \tau \) ... ... ... (i)
Thus,
The total time between the two meetings of the raft and the boat is \({t_1} + \tau = 2\tau \) ... ... ... (ii)
Now the total distance traveled by the raft is \(l\). This means that,
\(\left( {2\tau } \right) \times {V_r} = l\)
or, \({V_r} = \frac{l}{{2\tau }}\)
or, \({V_r} = \frac{l}{{2\tau }} = \frac{{6.0km}}{{2 \times 60\min }} = \frac{{6.0km}}{{2h}} = 3.0km/h\)