/*! 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 53 A swimmer, capable of swimming a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A swimmer, capable of swimming at a speed of \(1.4 \mathrm{m} / \mathrm{s}\) in still water (i.e., the swimmer can swim with a speed of \(1.4 \mathrm{m} / \mathrm{s}\) relative to the water), starts to swim directly across a \(2.8-\mathrm{km}\) -wide river. However, the current is \(0.91 \mathrm{m} / \mathrm{s},\) and it carries the swimmer downstream. (a) How long does it take the swimmer to cross the river? (b) How far downstream will the swimmer be upon reaching the other side of the river?

Short Answer

Expert verified
(a) 2000 seconds, (b) 1820 meters downstream.

Step by step solution

01

Identify Known Values

The speed of the swimmer in still water is given as \(v_s = 1.4\, \mathrm{m/s}\). The width of the river is \(2.8 \, \mathrm{km}\), which we convert to meters: \(2.8 \, \mathrm{km} = 2800 \, \mathrm{m}\). The current speed is \(v_c = 0.91 \, \mathrm{m/s}\).
02

Determine Time to Cross River

The time taken to swim across a river depends on the width of the river and the swimmer's speed perpendicular to the current. Since the swimmer swims directly across the river, the time \(t\) to cross is given by:\[t = \frac{\text{width of the river}}{v_s} = \frac{2800}{1.4} = 2000 \, \mathrm{s}\]
03

Calculate Downstream Distance

While the swimmer crosses the river, they are also carried downstream by the current. The downstream distance \(d\) that the swimmer is carried can be calculated using the current's speed and the time to cross:\[d = v_c \times t = 0.91 \, \mathrm{m/s} \times 2000 \, \mathrm{s} = 1820 \, \mathrm{m}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Relative Velocity
In river crossing problems, the concept of relative velocity plays a key role. Relative velocity helps us understand how different speeds interact in a moving medium like water.
When a swimmer moves through water that is itself moving, their effective velocity is a combination of their swimming speed and the speed of the river. This is because the swimmer is moving relative to the water, and the water is moving relative to the river banks.
  • If the swimmer is swimming perpendicular to the river flow, their speed remains effective across, but they get carried along by the current.
  • The total effective velocity of the swimmer involves both their own swimming speed and the speed of the river current.
The swimmer's velocity relative to the ground (or shore) can be found by vector addition of the swimmer's velocity relative to the water and the current's velocity. Understanding this helps in predicting where the swimmer will eventually reach once they cross the river.
Current Speed
The current speed of a river is another crucial factor to consider in such problems. It influences how far downstream a swimmer or an object will be carried during a crossing.
The speed of the current affects not just the crossing time but also the final position of the swimmer on the opposite shore. If a swimmer were to swim directly across without accounting for the current, they would end up downstream.
  • Current speed is usually measured in meters per second (m/s), and it varies greatly between different rivers.
  • In this problem, the current speed is given as 0.91 m/s, meaning the water is moving at this speed parallel to the river banks.
It's important to account for the current speed in distance calculations, as it alters trajectory and overall travel path.
River Width
River width is a straightforward yet essential concept in determining how long a crossing will take. It is simply the distance from one bank of the river to the opposite bank.
In problems like these, knowing the river width helps in calculating how long it will take a swimmer to reach the other side using their own swimming speed.
  • The width of the river is provided in this exercise as 2.8 kilometers, which, when converted to meters, is 2800 meters.
  • This distance remains constant and is considered the direct distance for the swimmer's crossing.
By knowing the swimmer's speed, the width can be used directly to find out how long it takes to swim across, assuming they swim perpendicular to the current.
Distance Calculation
Calculating distance in river crossing problems involves determining how far a swimmer or object is carried along the current while moving across.
Distance is a product of speed and time, and in this exercise, we calculate the downstream distance the swimmer is carried by multiplying the current speed by the time taken to cross the river.
  • The current moves the swimmer while they cross, allowing us to calculate the downstream slip.
  • For our swimmer, this involves multiplying 0.91 m/s (current speed) by 2000 seconds (time to cross). This calculation results in a downstream distance of 1820 meters.
This highlights how crucial it is to consider both river width for crossing time and current speed for overall distance traveled.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A skateboarder shoots off a ramp with a velocity of \(6.6 \mathrm{m} / \mathrm{s},\) directed at an angle of \(58^{\circ}\) above the horizontal. The end of the ramp is \(1.2 \mathrm{m}\) above the ground. Let the \(x\) axis be parallel to the ground, the \(+y\) direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

A volleyball is spiked so that it has an initial velocity of \(15 \mathrm{m} / \mathrm{s}\) directed downward at an angle of \(55^{\circ}\) below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?

A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of \(5.3 \mathrm{m} / \mathrm{s},\) hoping to land on the roof of an adjacent building. Air resistance is negligible. The horizontal distance between the two buildings is \(D,\) and the roof of the adjacent building is \(2.0 \mathrm{m}\) below the jumping-off point. Find the maximum value for \(D\).

A jetliner can fly 6.00 hours on a full load of fuel. Without any wind it flies at a speed of \(2.40 \times 10^{2} \mathrm{m} / \mathrm{s} .\) The plane is to make a round-trip by heading due west for a certain distance, turning around, and then heading due east for the return trip. During the entire flight, however, the plane encounters a \(57.8-\mathrm{m} / \mathrm{s}\) wind from the jet stream, which blows from west to east. What is the maximum distance that the plane can travel due west and just be able to return home?

The captain of a plane wishes to proceed due west. The cruising speed of the plane is \(245 \mathrm{m} / \mathrm{s}\) relative to the air. A weather report indicates that a \(38.0-\mathrm{m} / \mathrm{s}\) wind is blowing from the south to the north. In what direction, measured with respect to due west, should the pilot head the plane?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.