Problem 35 - Calculus: Early Transcendentals

Question

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 \( v = s'(t) \) and the acceleration is \( a = v'(t). \) If \( g \) is the acceleration due to gravity, them the downward force on the object is \( mg - cv, \) where \( c \) is a positive constant, and Newton's Second Law gives \( m \frac {dv}{dt} = mg - cv \) (a) Solve this as a linear equation to show that \( v = \frac {mg}{c} (1 - e^{-ct/m}) \) (b) What is the limiting velocity? (c) Find the distance the object has fallen after \( t \) seconds.

Step-by-Step Solution

Answer

(a) Solve to get \( v(t) = \frac{mg}{c}(1 - e^{-ct/m}) \). (b) Limiting velocity is \( \frac{mg}{c} \). (c) Distance is \( s(t) = \frac{mg}{c}(t + \frac{m}{c}(1 - e^{-ct/m})) \).

Step 1: Introduce the Differential Equation

We start with the given equation from Newton's Second Law: \( m \frac{dv}{dt} = mg - cv \) This represents the forces acting on the object: gravitational force \( mg \) and the opposing force due to air resistance \( cv \). We need to solve this differential equation to find the expression for velocity \( v(t) \).

Step 2: Rearrange the Equation

We rearrange the differential equation into standard linear form for solving linear ordinary differential equations: \( \frac{dv}{dt} + \frac{c}{m}v = g \) This reformatting helps us find an integrating factor to solve the equation.

Step 3: Find the Integrating Factor

The integrating factor \( \mu(t) \) is computed using the expression \( \mu(t) = e^{\int \frac{c}{m} dt} = e^{\frac{c}{m}t} \). Multiply through by the integrating factor to transform the equation into an exact differential: \( e^{\frac{c}{m}t} \frac{dv}{dt} + \frac{c}{m} e^{\frac{c}{m}t}v = g e^{\frac{c}{m}t} \).

Step 4: Solve the Transformed Equation

Now the left-hand side is the exact derivative of \( ve^{\frac{c}{m}t} \): \( \frac{d}{dt} \left( ve^{\frac{c}{m}t} \right) = g e^{\frac{c}{m}t} \). Integrate both sides with respect to \( t \): \( ve^{\frac{c}{m}t} = \frac{mg}{c} e^{\frac{c}{m}t} + C \), where \( C \) is the constant of integration.

Step 5: Solve for Velocity Expression

Solve for \( v(t) \) by dividing through by the integrating factor \( e^{\frac{c}{m}t} \): \( v(t) = \frac{mg}{c} - Ce^{-\frac{c}{m}t} \). Using the initial condition that the object is dropped from rest \( v(0) = 0 \), substitute into our equation, \( \frac{mg}{c} - C = 0 \), gives \( C = \frac{mg}{c} \). Thus, \( v(t) = \frac{mg}{c}(1 - e^{-\frac{c}{m}t}) \).

Step 6: Determine the Limiting Velocity

Taking the limit as \( t \to \infty \) of \( v(t) = \frac{mg}{c}(1 - e^{-\frac{c}{m}t}) \), the exponential term \( e^{-\frac{c}{m}t} \) goes to zero. So, the velocity approaches \( \frac{mg}{c} \), the limiting velocity.

Step 7: Find the Distance Fallen

The velocity function is \( v(t) = \frac{ds}{dt} = \frac{mg}{c}(1 - e^{-\frac{c}{m}t}) \). To find distance \( s(t) \), integrate: \( s(t) = \int \frac{mg}{c}(1 - e^{-\frac{c}{m}t}) dt = \frac{mg}{c}(t + \frac{m}{c}e^{-\frac{c}{m}t}) + D \). Substituting the initial condition \( s(0) = 0 \), find \( D = -\frac{mg}{c}\frac{m}{c} \). Therefore, \( s(t) = \frac{mg}{c} \left(t + \frac{m}{c}(1 - e^{-\frac{c}{m}t})\right) \).

Key Concepts

Newton's Second Law

Newton's Second Law is a fundamental principle in physics that describes the relation between a body's mass, its acceleration, and the applied force. The law is expressed with the formula \( F = ma \), where \( F \) denotes the force applied to an object, \( m \) is the mass, and \( a \) is the acceleration. This principle is crucial when analyzing motion. It provides a framework for understanding how objects respond to forces acting on them.

In the context of the given problem, Newton’s Second Law is applied to a falling object with air resistance as a force opposing the gravitational pull. Here, the equation \( m \frac{dv}{dt} = mg - cv \) captures the dynamic of both gravity (\( mg \)) and air resistance (\( cv \), where \( c \) is a proportionality constant) affecting the object's velocity over time.

This equation serves as the starting point for solving the velocity of the object as a function of time.

Limiting Velocity

The limiting velocity, also known as terminal velocity, is the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. In simpler words, it's the highest speed that the object attains when falling through a medium, such as air.

In our problem, limiting velocity is the point where the downward gravitational force is balanced by the upward force of air resistance. The distinction here is that when these forces equalize, acceleration stops, and the object continues to fall at a constant speed. Mathematically, this is shown by taking the limit of the velocity function \( v(t) = \frac{mg}{c}(1 - e^{-\frac{ct}{m}}) \) as \( t \to \infty \).

As \( t \to \infty \), the exponential component \( e^{-\frac{ct}{m}} \) approaches zero, and thus, the velocity achieves the value \( \frac{mg}{c} \). That's the limiting velocity of the falling object.

Integrating Factor

An integrating factor is a crucial technique used in solving linear ordinary differential equations, especially when dealing with first-order linear equations. This method simplifies these equations into a form that is easier to integrate directly.

In our specific case, after rearranging the differential equation to \( \frac{dv}{dt} + \frac{c}{m}v = g \), we identify the standard linear form. The integrating factor, \( \mu(t) \), is calculated as \( e^{\int \frac{c}{m} dt} = e^{\frac{ct}{m}} \). This factor is multiplied through the equation to convert it into an exact derivative, allowing easy integration.

The beauty of using an integrating factor lies in its ability to convert a complex differential equation into a straightforward one. It simplifies the solution process and helps derive the explicit form of the velocity function for the object.

Velocity Function

The velocity function describes how the velocity of an object changes over time. It is derived by integrating the differential equation obtained from Newton's Second Law.

For this exercise, the velocity function is given as \( v(t) = \frac{mg}{c}(1 - e^{-\frac{ct}{m}}) \). This represents the speed of the object at any time \( t \), factoring in initial conditions. At \( t = 0 \), the object starts from rest, confirming our boundary condition that helps find the constant of integration.

This equation holds significant meaning: initially, the velocity increases as the object accelerates under gravity. As time increases, the rate of change of velocity diminishes until it reaches the limiting velocity, \( \frac{mg}{c} \), due to air resistance slowing it down. The velocity function, thus, provides comprehensive insight into the object's motion through time.