/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 You are a member of a geological... [FREE SOLUTION] | 91Ó°ÊÓ

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You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determine the width of the river and the current speed (the speed of the water relative to the earth). You have a small boat with an outboard motor. By measuring the time it takes to cross a pond where the water isn't flowing, you have calibrated the throttle settings to the speed of the boat in still water. You set the throttle so that the speed of the boat relative to the river is a constant \(6.00 \mathrm{~m} / \mathrm{s}\). Traveling due north across the river, you reach the opposite bank in \(20.1 \mathrm{~s}\). For the return trip, you change the throttle setting so that the speed of the boat relative to the water is \(9.00 \mathrm{~m} / \mathrm{s}\). You travel due south from one bank to the other and cross the river in \(11.2 \mathrm{~s}\). (a) How wide is the river, and what is the current speed? (b) With the throttle set so that the speed of the boat relative to the water is \(6.00 \mathrm{~m} / \mathrm{s},\) what is the shortest time in which you could cross the river, and where on the far bank would you land?

Short Answer

Expert verified
The width of the river is \(120.6 \, m\) and the current speed is \(1.79 \, m/s\). The shortest time to cross the river (with a throttle setting having speed of the boat relative to water at \(6.00 \, m/s\)) is \(21.5 \, s\), and the landing point on the far bank will be \(38.5 \, m\) downstream.

Step by step solution

01

Calculate the width of the river

The boat's speed relative to the river is given as \(6.00 \, m/s\), and the time taken to cross the river is \(20.1 \, s\). The width (distance) of the river can be calculated using the equation speed = distance/time. Therefore, the width of the river is \(6.00 \, m/s \times 20.1 \, s = 120.6 \, m\).
02

Compute the current speed

To calculate the current speed, we use the information from the southward journey. The speed of the boat relative to the earth is the vector sum of the speed of the water and the speed of the boat relative to the water. This gives us the equation \(v_{\text{be}} = v_{\text{bw}} + v_{\text{we}}\), where \(v_{\text{be}}\) is the speed of the boat relative to the earth, \(v_{\text{bw}}\) is the speed of the boat relative to the water, and \(v_{\text{we}}\) is the speed of the water relative to the earth (or the current speed). From the southward journey, we have \(v_{\text{be}} = \text{width of the river} / \text{time}\), and \(v_{\text{bw}}\) is given as \(9.00 \, m/s\). So, the current speed can be solved by \(v_{\text{we}} = v_{\text{be}} - v_{\text{bw}} = 120.6 \, m / 11.2 \, s - 9.00 \, m/s = 10.79 \, m/s - 9.00 \, m/s = 1.79 \, m/s\).
03

Determine the shortest time to cross and landing point on the far bank

With a speed setting above the current speed (\(6.00 \, m/s > 1.79 \, m/s\)), the shortest time will be achieved by steering the boat perpendicular to the current. The speed of the boat relative to earth, under vector decomposition, will be less than the speed of the boat relative to water, and is given by \(v_{\text{be}} = \sqrt{v_{\text{bw}}^2 - v_{\text{we}}^2} = \sqrt{(6.00 \, m/s)^2 - (1.79 \, m/s)^2} = 5.60 \, m/s\). The shortest time to cross will then be \(\text{width of the river} / v_{\text{be}} = 120.6 \, m / 5.60 \, m/s = 21.5 \, s\). The point of landing on the far bank will be downstream due to the current, which can be calculated by \(v_{\text{we}} \times \text{time}\) = \(1.79 \, m/s \times 21.5 \, s = 38.5 \, m\).

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Key Concepts

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

Vector Addition
Vector addition is an essential mathematical tool used in river crossing physics to determine combined velocities. Imagine vectors as arrows in space, each pointing in a different direction with its length representing speed. When crossing a river in a boat, both the boat speed and the river current act as different vectors. Their resultant performance needs to be assessed. The speed relative to the earth is a vector sum of both these influences. Mathematically, if one vector represents the river's flow and the other the boat's speed relative to the water, we find the net velocity by adding these vectors head-to-tail. Understanding vector addition is crucial for accurate navigation across a river where both the boat's velocity and the current's velocity must be considered.
Relative Motion
Relative motion helps us understand how different observers perceive the velocity of moving objects. In the context of boat navigation across a river, it explains how different velocities appear when viewed relative to different frames of reference. Here, we have three velocities:
  • The boat's speed relative to the water ( \( v_{ ext{bw}} \) ).
  • The water's speed relative to the earth, known as the current speed ( \( v_{ ext{we}} \) ).
  • The boat's speed relative to the earth ( \( v_{ ext{be}} \) ), resulting from the combination of the above two.
Using relative motion, we establish that \( v_{ ext{be}} = v_{ ext{bw}} + v_{ ext{we}} \) during crossing, highlighting the importance of considering all components while assessing real-world movements and timings across the river.
Boat Navigation
Navigating a boat across a river requires understanding both the boat's capabilities and environmental factors such as current speed. Successful crossing involves planning a course that accounts for the current, which can push the boat off the intended path.
For straight-line navigation seeking the shortest crossing time, you aim perpendicular to the current while adjusting power settings to maintain control. If the goal is to land directly opposite the starting point, you must compensate for the river's current by angling the boat upstream. Using motor speed and direction to counteract the current allows for precise control over the eventual landing point.
Current Speed Calculation
Calculating the current speed involves analyzing how the water movement alters the boat's movement from its path in still water. Usually, you derive it from the boat's altered speed during travel.
In this scenario, you first establish the river width from the boat's transit time when steering due north. By comparing the boat's velocity relative to the river to its velocity relative to the earth, you can infer the current speed. It involves understanding the change between these speeds using vector mathematics: the boat's forward velocity minus the resultant velocity when considering directional deviations gives you the current's speed.The formula \( v_{ ext{we}} = v_{ ext{be}} - v_{ ext{bw}} \) effectively computes the current's influence, enlightening broader navigation techniques on flowing water bodies.

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Most popular questions from this chapter

The radius of the earth's orbit around the sun (assumed to be circular) is \(1.50 \times 10^{8} \mathrm{~km},\) and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in \(\mathrm{m} / \mathrm{s} ?(\mathrm{~b})\) What is the magnitude of the radial acceleration of the earth toward the sun, in \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius \(=5.79 \times 10^{7} \mathrm{~km},\) orbital period \(=88.0\) days).

An airplane pilot sets a compass course due west and maintains an airspeed of \(220 \mathrm{~km} / \mathrm{h}\). After flying for \(0.500 \mathrm{~h},\) she finds herself over a town \(120 \mathrm{~km}\) west and \(20 \mathrm{~km}\) south of her starting point. (a) Find the wind velocity (magnitude and direction). (b) If the wind velocity is \(40 \mathrm{~km} / \mathrm{h}\) due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of \(220 \mathrm{~km} / \mathrm{h}\).

A toy rocket is launched with an initial velocity of \(12.0 \mathrm{~m} / \mathrm{s}\) in the horizontal direction from the roof of a \(30.0-\mathrm{m}\) -tall building. The rocket's engine produces a horizontal acceleration of \(\left(1.60 \mathrm{~m} / \mathrm{s}^{3}\right) t,\) in the same direction as the initial velocity, but in the vertical direction the acceleration is \(g\), downward. Ignore air resistance. What horizontal distance does the rocket travel before reaching the ground?

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A man stands on the roof of a 15.0 -m-tall building and throws a rock with a speed of \(30.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(33.0^{\circ}\) above the horizontal. Ignore air resistance. Calculate (a) the maximum height above the roof that the rock reaches; (b) the speed of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. (d) Draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

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