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Chapter 5: Problem 26

A river is flowing from west to east at a speed of \(5 \mathrm{mmin}^{-1}\). A man on the south bank of the river, capable of swimming at \(10 \mathrm{~m} \min ^{-1}\) in still water, wants to swim across the river in the shortest time. Finally, he will move in a direction (1) \(\tan ^{-1}(2) \mathrm{E}\) of \(\mathrm{N}\) (2) \(\tan ^{-1}(2) \mathrm{N}\) of \(\mathrm{E}\) (3) \(30^{\circ} \mathrm{E}\) of \(\mathrm{N}\) (4) \(60^{\circ} \mathrm{E}\) of \(\mathrm{N}\)

Short Answer

Expert verified
The man should swim at a direction \(30^{\circ} \mathrm{E}\) of \(\mathrm{N}\).

Step by step solution

01

Understanding the Problem

The man wants to cross the river in the shortest time possible. The river flows at a speed of \(5 \mathrm{~m} \min^{-1}\) from west to east. The man can swim at \(10 \mathrm{~m} \min^{-1}\) in still water. We need to find the angle at which the man should swim to achieve this.
02

Identify Velocity Components

To swim across quickest, the swimmer should have a velocity perpendicular to the river's flow. This means his velocity component towards the north should be maximized.
03

Determine Perpendicular Velocity Component

The swimmer's resultant velocity should have a component of his swimming speed directed against the river's eastward flow such that his northward component is maximized. If \(\theta\) is the angle with the north direction, \(10\sin(\theta)\) should counteract the river current: \(10\sin(\theta) = 5\).
04

Solve for the Angle

Solving \(\sin(\theta) = 0.5\), we find \(\theta = 30^\circ\).
05

Movement Direction

The man should swim in a direction \(30^\circ\) east of north to have the shortest crossing time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

River Crossing Problem
A river crossing problem in physics typically involves understanding how to navigate across a flowing body of water efficiently. Imagine you're standing on a riverbank, and you want to cross to the opposite side as quickly as possible.
This isn't just about swimming straight from point to point. Since rivers flow, their current can push swimmers away from their intended path. This means we must consider both the swimmer's speed and the river's current to find the quickest crossing route.
In our exercise, the river flows from west to east. To cross swiftly, the swimmer must overcome this eastward flow. To do that, he needs to swim at a specific angle relative to the riverbank to counter the current's push. This will allow him to reach the opposite bank in the shortest possible time.
  • River's Current: 5 meters per minute eastward.
  • Swimmer鈥檚 Speed: 10 meters per minute in still water.
Recognizing this is the first step in solving any river crossing problem.
Velocity Components
Velocity components help us understand the different parts of a bustling speed. In river crossing problems, we examine how a swimmer's velocity can be split into two parts: one that works against the river's flow and another that moves straight across the river.
Imagine slicing the swimmer鈥檚 speed into a horizontal and vertical component. The horizontal component fights against the river鈥檚 motion, while the vertical component pushes him across to the other side.
The objective is to maximize the northward, or perpendicular, component of velocity, ensuring a swift crossing. In doing so, the swimmer balances counteracting the eastward flow, effectively using energy to propel more northward than eastward.
  • East-West Component: Opposes the river's current.
  • North-South Component: Moves the swimmer across the river.
By understanding these components, we can discover the best angle for swimming.
Trigonometry in Physics
Trigonometry is like a close friend to physics, especially in problems dealing with angles and directions, such as river crossings. It helps us to relate the angle chosen by the swimmer to the velocity components needed to tackle the river's current.
Using trigonometry, we can break down a swimmer鈥檚 velocity into parts that describe northward and eastward movements. Our key equation comes from knowing that \(\sin(\theta)\) describes the ratio of the opposite side to the hypotenuse in a right triangle 鈥� effectively our swimmer鈥檚 strategy for countering the current while swimming straight across.
Given: \(10 \sin(\theta) = 5\), we solve for \(\theta\) by rearranging to determine that \(\sin(\theta) = 0.5\). From trigonometric tables or calculations, we find that \(\theta = 30^\circ\). This angle allows him to strike the perfect balance between countering the eastward push and his northward swimming target.
This math helps us solve practical problems, giving real-world significance to geometric principles.

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Most popular questions from this chapter

Shots are fired simultaneously from the top and bottom of a vertical cliff with the elevation \(\alpha=30^{\circ}\), \(\beta=60^{\circ}\), respectively (figure). The shots strike an object simultaneously at the same point. If \(a=30 \sqrt{3} \mathrm{~m}\) is the horizontal distance of the object from the cliff, then the height \(h\) of the cliff is (1) \(30 \mathrm{~m}\) (2) \(45 \mathrm{~m}\) (3) \(60 \mathrm{~m}\) (4) \(90 \mathrm{~m}\)

A swimmer wishes to cross a \(500-\mathrm{m}\) river flowing at \(5 \mathrm{~km} \mathrm{~h}^{-1}\). His speed with respect to water is \(3 \mathrm{~km} \mathrm{~h}^{-1}\). The shortest possible time to cross the river is (1) \(10 \mathrm{~min}\) (2) \(20 \mathrm{~min}\) (3) \(6 \mathrm{~min}\) (4) \(7.5 \mathrm{~min}\)

Rain appears to fall vertically to a man walking at \(3 \mathrm{~km} \mathrm{~h}^{-1}\), but when he changes his speed to double, the rain appears to fall at \(45^{\circ}\) with vertical. Study the following statements and find which of them are correct. i. Velocity of rain is \(2 \sqrt{3} \mathrm{~km} \mathrm{~h}^{-1}\) ii. The angle of fall of rain (with vertical) is $$ \theta=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right) \text {. } $$ iii. The angle of fall of rain (with vertical) is $$ \theta=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) \text {. } $$ iv. Velocity of rain is \(3 \sqrt{2} \mathrm{kmh}^{-1}\). (I) Statements (i) and (ii) are correct. (2) Statements (i) and (iii) are correct. (3) Statements (iii) and (iv) are correct. (4) Statements (ii) and (iv) are correct.

A body is projected with velocity \(u\) at an angle of projection \(\theta\) with the horizontal. The direction of velocity of the body makes angle \(30^{\circ}\) with the horizontal at \(t=2 \mathrm{~s}\) and then after \(1 \mathrm{~s}\) it reaches the maximum height. Then (1) \(u=20 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\) (2) \(\theta=60^{\circ}\) (3) \(\theta=30^{\circ}\) (4) \(u=10 \sqrt{3} \mathrm{~ms}^{-1}\)

Points \(A\) and \(C\) are on the horizontal ground and \(A\) and \(B\) are in same vertical plane at a distance of \(1500 \mathrm{~m}\). Simultaneously bullets are fired from \(A, B\) and \(C\) and they collide at \(O\). The bullet at \(B\) is fired at an angle of \(30^{\circ}\) with horizontal towards the ground at velocity \(100 \mathrm{~m} / \mathrm{s}\). The bullet at \(C\) is projected vertically upward at velocity of \(100 \mathrm{~m} / \mathrm{s}\). The bullet projected from \(A\) reaches its maximum height at \(O\). Find the velocity of bullet at \(A\) : (1) \(50 \mathrm{~m} / \mathrm{s}\) (2) \(50 \sqrt{7} \mathrm{~m} / \mathrm{s}\) (3) \(60 \sqrt{7} \mathrm{~m} / \mathrm{s}\) (4) \(60 \mathrm{~m} / \mathrm{s}\)
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