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Chapter 2: Problem 121

A river is flowing from west to east with a speed of \(5 \mathrm{~m} / \mathrm{min}\). A man can swim in still water with a velocity \(10 \mathrm{~m} / \mathrm{min}\). In which direction should the man swim, so as to take the shortest possible path to go to the south? (A) \(30^{\circ}\) with downstream (B) \(60^{\circ}\) with downstream (C) \(120^{\circ}\) with downstream (D) Towards south

Short Answer

Expert verified
The man should swim in a direction making an angle of \(60^{\circ}\) with the downstream direction to take the shortest possible path to the south. So, the correct answer is (B) \(60^{\circ}\) with downstream.

Step by step solution

01

Analyzing the situation

First, we draw a triangle to represent the three velocities involved in this problem: the velocity of the man in river frame (10m/min), the velocity of the river current (5m/min), and the actual velocity of the man in the ground frame. If we consider the eastward direction positive, we represent the velocities as follows: - Let the velocity of the man in still water be \(\vec{v_m} = 10\mathrm{~m/min}\) - Let the velocity of the river current be \(\vec{v_r} = 5\mathrm{~m/min}\) - Let the man's actual velocity w.r.t the ground be \(\vec{v}_{a}\)
02

Setting up the vectors

We have three vectors, where the actual velocity of the man is the vector sum of the river's current velocity and the man's velocity in still water. If the man wants to swim directly south, we can write the vector mathematically as: \[\vec{v}_{a} = \vec{v_m} + \vec{v_r}\] Now if we let \(\theta\) be the angle between the man's actual velocity and the downstream direction, we can write the man's velocity in still water as: \[\vec{v_m} = 10\mathrm{~m/min}\times(cos(\theta)\hat{i}+ sin(\theta)\hat{j})\]
03

Finding the shortest path

The shortest path to go to the south is when the man swims in such a way that the eastward component of his velocity just cancels the westward current of the river so that the only remaining component of his velocity is in the south direction. So, we set up the equation relating the velocity components: \[5\mathrm{~m/min} = 10\mathrm{~m/min}\times cos(\theta)\]
04

Solving for the angle

Solving the equation for the angle \(\theta\): \[cos(\theta) = \frac{5\mathrm{~m/min}}{10\mathrm{~m/min}} = \frac{1}{2}\] \[\theta = cos^{-1}\left(\frac{1}{2}\right) = 60^{\circ}\] Hence, the man should swim in a direction making an angle of \(60^{\circ}\) with the downstream direction to take the shortest possible path to the south. So, the correct answer is (B) \(60^{\circ}\) with downstream.

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Most popular questions from this chapter

A cricketer can throw a ball to a maximum horizontal distance of \(100 \mathrm{~m}\). How much high above the ground can the cricketer throw the same ball? (A) \(50 \mathrm{~m}\) (B) \(100 \mathrm{~m}\) (C) \(150 \mathrm{~m}\) (D) \(200 \mathrm{~m}\)

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A man wants to reach point \(C\) on the opposite bank of river. He can swim in still water with speed \(5 \mathrm{~m} / \mathrm{s}\) and can walk on ground with \(10 \mathrm{~m} / \mathrm{s}\). Velocity of river is \(3 \mathrm{~m} / \mathrm{s}\) and given \(A B=B C=500 \mathrm{~m} .\) $$ \begin{aligned} &\text { Column-I } & \text { Column-II } \\ &\hline \text { (A) The time to reach point } C \text { if } & \text { (1) } 125 \mathrm{~s} \\ &\text { he crosses the river by shortest } \\ &\text { path is } \\ &\text { (B) The time to reach point } C \text { if } & \text { (2) } 175 \mathrm{~s} \\ &\text { he crosses the river in shortest } \\ &\text { time. } \\ &\text { (C) The time to reach point } C \text { if } & \text { (3) } 120 \mathrm{~s} \\ &\text { he crosses the river along } & \text { (4) } 106 \mathrm{~s} \\ &\text { path } A C \text { (approximate value). } & \\ &\text { (D) Time to reach point } B \\ &\text { with zero drift } \end{aligned} $$
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