/*! 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 35 An object with mass \(m\) is dro... [FREE SOLUTION] | 91Ó°ÊÓ

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An object with mass \(m\) is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If \(s(t)\) is the distance dropped after \(t\) seconds, then the speed is \(v=s^{\prime}(t)\) and the acceleration is \(a=v^{\prime}(t) .\) If \(g\) is the acceleration due to gravity, then the downward force on the object is \(m g-c v,\) where \(c\) is a positive constant, and Newton's Second Law gives $$ m \frac{d v}{d t}=m g-c v $$ (a) Solve this as a linear equation to show that $$ v=\frac{m g}{c}\left(1-e^{-c t / m}\right) $$ (b) What is the limiting velocity? (c) Find the distance the object has fallen after \(t\) seconds.

Short Answer

Expert verified
(a) Linear solution: \( v = \frac{mg}{c} (1 - e^{-ct/m}) \). (b) Limiting velocity: \( \frac{mg}{c} \). (c) Distance: \( s(t) = \frac{mg}{c} t + \frac{m^2 g}{c^2} (1 - e^{-ct/m}) \).

Step by step solution

01

Identify the Type of Equation

The equation provided is a first-order linear differential equation: \( m \frac{dv}{dt} = mg - cv \). This equation represents the rate of change of velocity with respect to time.
02

Separating Variables

Rearrange the differential equation: \( m \frac{dv}{dt} + cv = mg \). Identify the homogeneous part: \( \frac{dv}{dt} + \frac{c}{m}v = g \). This form allows us to proceed with finding an integrating factor.
03

Find Integrating Factor

The integrating factor \( \mu(t) \) is \( e^{\int \frac{c}{m} dt} = e^{\frac{c}{m} t} \). Multiply through the differential equation by this factor: \( e^{\frac{c}{m} t} \frac{dv}{dt} + \frac{c}{m}e^{\frac{c}{m} t}v = ge^{\frac{c}{m} t} \).
04

Solve Differential Equation

Recognize the left hand side as the derivative of a product: \( \frac{d}{dt}(e^{\frac{c}{m}t} v) = ge^{\frac{c}{m} t} \). Integrate both sides: \( e^{\frac{c}{m} t} v = \int ge^{\frac{c}{m} t} dt = \frac{mg}{c}e^{\frac{c}{m} t} + C \).
05

Solve for v

Solve the equation for \( v \): \( v = \frac{mg}{c} - Ce^{-\frac{c}{m} t} \). Use initial condition (\( v(0) = 0 \)) to find \( C \): \( C = \frac{mg}{c} \). Thus, \( v = \frac{mg}{c}(1-e^{-\frac{c}{m}t}) \).
06

Find Limiting Velocity

The limiting velocity \( v_l \) is the value of \( v \) as \( t \) approaches infinity. Calculate \( v_l = \lim_{t \to \infty} \frac{mg}{c}(1-e^{-\frac{c}{m}t}) = \frac{mg}{c} \) since \( e^{-\frac{c}{m}t} \to 0 \).
07

Integrate to Find Distance

Find distance \( s(t) \) by integrating \( v(t) \): \( s(t) = \int \frac{mg}{c}(1-e^{-\frac{c}{m}t}) dt \). This yields \( s(t) = \frac{mg}{c}t + \frac{m^2g}{c^2}e^{-\frac{c}{m} t} + D \). Use initial condition \( s(0) = 0 \) to find \( D = -\frac{m^2g}{c^2} \). Thus, \( s(t) = \frac{mg}{c}t + \frac{m^2g}{c^2}(1-e^{-\frac{c}{m} t}) \).

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Key Concepts

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

Linear differential equation
A linear differential equation is a type of differential equation that can be written in the form: \( \frac{dy}{dt} + p(t)y = q(t) \). In the context of this exercise, we deal with such an equation: \( m \frac{dv}{dt} = mg - cv \). This equation describes how the velocity \( v \) of an object changes over time \( t \). Here, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( c \) is a constant representing the proportionality to the velocity, often due to air resistance.
By converting the equation to: \( \frac{dv}{dt} + \frac{c}{m}v = g \), it sets up a first-order linear differential equation suitable for solving with the method of integrating factors. The homogeneous structure and linear terms make it possible to solve for velocity using simple integration methods and techniques.
Newton's Second Law
Newton's Second Law is a foundational principle in physics that explains how the velocity of an object changes when it is subjected to external forces. It is commonly expressed as \( F = ma \), where \( F \) is the net force acting on an object, \( m \) is the object's mass, and \( a \) is its acceleration.
In this exercise, the law is applied to an object in free fall with air resistance. The equation \( m \frac{dv}{dt} = mg - cv \) expresses the forces acting on the object:
  • \( mg \) is the force due to gravity, pulling it downwards.
  • \( cv \) is the force of air resistance opposing the motion.
The result is a dynamic balance between these two forces, dictating how the object's velocity changes as it accelerates towards its limiting velocity.
Integrating factor
An integrating factor is a function used to simplify the solving process of linear differential equations. It is typically represented as \( \mu(t) = e^{\int p(t)dt} \), which helps in converting a non-exact differential equation into an exact one, enabling straightforward integration.
In our exercise, the integrating factor is derived as \( e^{\frac{c}{m} t} \). By multiplying the entire differential equation by this factor, we prepare it for direct integration. This transforms the left-hand side into a single derivative term, making it integrable:
  • \( e^{\frac{c}{m} t} \frac{dv}{dt} + \frac{c}{m}e^{\frac{c}{m} t}v = ge^{\frac{c}{m} t} \).
Integrating both sides gives us the expression that leads to the solution for velocity, particularly beneficial in solving physics problems like motion with resistance.
Limiting velocity
Limiting velocity, often referred to as "terminal velocity," is the constant speed that a freely falling object eventually reaches when the force due to gravity is balanced by the drag force due to air resistance.
In this scenario, the limiting velocity is found by observing the expression for velocity \( v = \frac{mg}{c}(1-e^{-\frac{c}{m}t}) \) as \( t \rightarrow \infty \). As time progresses, the exponential term \( e^{-\frac{c}{m}t} \rightarrow 0 \), simplifying the equation to:
  • \( v = \frac{mg}{c} \)
This signifies that the object continues to fall at this constant speed without further acceleration or deceleration, a very important concept in understanding dynamic motion in physics.

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