/*! 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 32 Find the equilibrium solution to... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 32

Find the equilibrium solution to \(d v / d t=\) \(-g-(c / m) v^{2}\). What is the limiting velocity?

Short Answer

Expert verified
The limiting velocity is \( v = \text{sqrt} \frac{mg}{c} \.

Step by step solution

01

- Understand the problem

To find the equilibrium solution to the differential equation \(\frac{d v}{d t} = -g - \frac{c}{m} v^{2}\), identify the conditions required for equilibrium.
02

- Set the derivative to zero

For equilibrium, the velocity must be constant, so set \( \frac{d v}{d t} = 0 \). This gives: \ 0 = -g - \frac{c}{m} v^{2} \.
03

- Solve for v

Rearrange the equation to solve for \( v \): \ \frac{c}{m} v^{2} = -g \. Then, multiply both sides by \( -1 \) to get: \ \frac{c}{m} v^{2} = g \.
04

- Isolate \( v \)

Isolate \( v \) by multiplying both sides by \( \frac{m}{c} \): \ v^{2} = \frac{mg}{c} \.
05

- Take the square root

Take the square root of both sides to solve for \( v \): \ v = \frac{\text{sqrt}(mg)}{\text{sqrt}(c)} \ or \ v = \text{sqrt} \frac{mg}{c}\.
06

- Identify the limiting velocity

The limiting velocity is the equilibrium solution, which is the value of \( v \) found in the previous step.

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 91影视!

Key Concepts

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

limiting velocity
Limiting velocity is the maximum speed an object can reach when considering forces like gravity and air resistance. When an object falls, gravity pulls it downward. But air resistance opposes this motion which increases with speed. Eventually, the forces balance out, leading to a constant velocity. This is known as the limiting or terminal velocity. The velocity stops increasing as the downward force of gravity is equal to the upward force of air resistance. Hence, an equilibrium is reached. In our problem, the limiting velocity is found by solving the equation when the acceleration is zero. This gives us \( v = \text{sqrt} \frac{mg}{c}\).
constant velocity
Constant velocity means an object's speed and direction do not change over time. This implies no acceleration. In our equation, constant velocity is achieved when the derivative of velocity with respect to time \( \frac{d v}{d t} = 0 \). At this point, forces acting on the object are balanced. In practical scenarios like free fall with air resistance, reaching constant velocity means the object has hit its limiting velocity. For differential equations, setting the rate of change to zero is a common method to identify steady-state or equilibrium solutions. Thus, constant velocity is a state where there is neither speeding up nor slowing down.
solving differential equations
Solving differential equations involves finding a function that satisfies the equation's condition. In our case, we start from the differential equation \(\frac{d v}{d t} = -g - \frac{c}{m} v^{2}\). First, determine the equilibrium state where \(\frac{d v}{d t} = 0\). This simplifies to \(0 = -g - \frac{c}{m} v^{2}\). We solve for \(v\) to find the steady-state solution. By isolating \(v\) and taking the square root, we get \(v = \text{sqrt} \frac{mg}{c}\). This method is typical for first-order ordinary differential equations. Equilibrium solutions often reveal important behavior about the system's dynamics over time.

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

Suppose that mold grows at a rate proportional to the amount present. If there are initially \(500 \mathrm{~g}\) of mold and \(6 \mathrm{~h}\) later there are \(600 \mathrm{~g}\), determine the amount of mold present after one day. When is the amount of mold \(1000 \mathrm{~g}\) ?

Determine the time of death if the temperature of a corpse is \(79^{\circ} \mathrm{F}\) when discovered at \(3: 00\) p.m. and \(68^{\circ} \mathrm{F} 3\) h later. Assume that the temperature of the surroundings is \(60^{\circ} \mathrm{F}\) (normal body temperature is \(98.6^{\circ} \mathrm{F}\) ).

Solve the Logistic equation, \(d y / d t=\alpha y(1-\) \((1 / K) y\) ), by viewing it as a Bernoulli equation and then solve the resulting linear equation by using an integrating factor rather than the method of undetermined coefficients that is illustrated in the examples.

Consider a solution to the logistic equation with initial population \(y_{0}\) where \(0

(Escape Velocity) Suppose that a rocket is launched from the Earth's surface. At a great (radial) distance \(r\) from the center of the Earth, the rocket's acceleration is not the constant \(g\). Instead, according to Newton's law of gravitation, \(a=k / r^{2}\), where \(k\) is the constant of proportionality ( \(k>0\) if the rocket is falling toward the Earth; \(k<0\) if the rocket is moving away from the Earth). (a) If \(a=-g\) at the Earth's surface (when \(r=R\) ), find \(k\) and show that the rocket's velocity is found by solving the initial value problem \(d v / d t=-g R^{2} / r^{2}, v(0)=v_{0}\). (b) Show that \(d v / d t=v d v / d r\) so that the solution to the initial value problem \(v d v / d r=\) \(-g R^{2} / r^{2}, v(R)=v_{0}\) is \(v^{2}=2 g R^{2} / r+v_{0}^{2}-\) \(2 g R\). (c) If \(v>0\) (so that the rocket does not fall to the ground), show that the minimum value of \(v_{0}\) for which this is true (even for very large values of \(r\) ) is \(v_{0}=\sqrt{2 g R}\). This value is called the escape velocity and signifies the minimum velocity required so that the rocket does not return to the Earth.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation
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.