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Chapter 9: Problem 46

An object of mass \(m\) is moving horizontally through a medium which resists the motion with a force that is a function of the velocity; that is, $$m \frac{d^{2} s}{d t^{2}}=m \frac{d v}{d t}=f(v)$$ where \(v=v(t)\) and \(s=s(t)\) represent the velocity and position of the object at time \(t,\) respectively. For example, think of a boat moving through the water. (a) Suppose that the resisting force is proportional to the velocity, that is, \(f(v)=-k v, k\) a positive constant. (This model is appropriate for small values of \(v . )\) Let \(v(0)=v_{0}\) and \(s(0)=s_{0}\) be the initial values of \(v\) and \(s\) Determine \(v\) and \(s\) at any time \(t .\) What is the total distance that the object travels from time \(t=0 ?\)

Short Answer

Expert verified
The velocity is \( v(t) = v_0 e^{-\frac{k}{m}t} \) and position is \( s(t) = -\frac{mv_0}{k} e^{-\frac{k}{m}t} + s_0 + \frac{mv_0}{k} \). Total distance is \( \frac{mv_0}{k} \).

Step by step solution

01

Write the given differential equation

We have the differential equation given by the resisting force formula: \( m \frac{dv}{dt} = -kv \). The right side represents the resistance force as a function of velocity.
02

Rearrange and separate variables

Rearrange the equation to isolate terms containing \(v\) and \(t\) on opposite sides: \( \frac{dv}{v} = -\frac{k}{m} dt \). This equation can be solved by integrating both sides.
03

Integrate both sides

Integrate both sides of the equation: \( \int \frac{dv}{v} = \int -\frac{k}{m} dt \). Thus, we get \( \ln |v| = -\frac{k}{m} t + C_1 \), where \(C_1\) is an integration constant.
04

Solve for the constant of integration

Use the initial condition \(v(0) = v_0\) to solve for \(C_1\). Substituting the initial conditions, \( \ln |v_0| = C_1 \). Therefore, the equation becomes \( \ln |v| = -\frac{k}{m}t + \ln |v_0| \).
05

Solve for velocity function

Exponentiate both sides to solve for \(v(t)\): \( v(t) = v_0 e^{-\frac{k}{m}t} \). This gives the velocity of the object as a function of time.
06

Integrate velocity to find position

The position function can be found by integrating the velocity function: \( s(t) = \int v(t) dt = \int v_0 e^{-\frac{k}{m}t} dt \). This yields \( s(t) = -\frac{mv_0}{k} e^{-\frac{k}{m}t} + C_2 \).
07

Solve for position constant

Use the initial condition \(s(0) = s_0\) to find \(C_2\): By substituting \(t = 0\), \( s_0 = -\frac{mv_0}{k} + C_2 \), thus \( C_2 = s_0 + \frac{mv_0}{k} \). The position function is then \( s(t) = -\frac{mv_0}{k} e^{-\frac{k}{m}t} + \left(s_0 + \frac{mv_0}{k}\right) \).
08

Determine the total distance traveled

As \(t \to \infty\), the velocity \(v(t)\) approaches zero, meaning the object stops. The total distance can be found as the limit of \(s(t)\) as \(t \to \infty\), resulting in \( s_\infty = s_0 + \frac{mv_0}{k} \). The total distance traveled is \( \frac{mv_0}{k} \).

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Key Concepts

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

Motion in Resisting Medium
When an object moves through a resisting medium like water or air, it experiences a force opposing its motion. This force depends on the velocity of the object. For small velocities, it is often assumed that the resisting force is proportional to velocity. In mathematical terms, this can be represented as \( f(v) = -k v \), where \( k \) is a constant indicating the strength of the resistive force.
The negative sign shows that the force acts in the opposite direction of motion. This model is used to describe scenarios such as a small boat moving through water, where friction and drag slow the boat down. Understanding this concept gives us insights into the dynamics of objects influenced by similar forces in various fields, from engineering to natural science.
It鈥檚 fascinating to note that as the resisting force slows down the object over time, the velocity gradually reduces to zero, and the movement ceases completely when only resistive forces are acting.
Separation of Variables
To solve differential equations like the one encountered in resisting motion, one common technique used is **Separation of Variables**. This method involves rearranging the equation such that all terms involving one variable (like velocity \( v \)) are on one side of the equation, and all terms involving another variable (like time \( t \)) are on the other.
In our example, we begin with the equation \( m \frac{dv}{dt} = -kv \). By rearranging, we isolate the differentials: \( \frac{dv}{v} = -\frac{k}{m} dt \).
This format allows us to individually integrate each side of the equation, a critical step to finding solutions. Separation of variables is an invaluable tool in solving first-order differential equations, frequently employed in physics and engineering problems where relationships between different quantities must be determined.
Integration Techniques
Integration is a fundamental process used to solve differential equations after separating variables. The goal is to find expressions for the dependent variable. For instance, to determine the velocity function \( v(t) \) from the separated equation \( \frac{dv}{v} = -\frac{k}{m} dt \), each side must be integrated separately.
  • Integrating \( \frac{dv}{v} \) results in \( \ln |v| \).
  • Integrating \( -\frac{k}{m} dt \) gives \( -\frac{k}{m} t + C \) where \( C \) is the integration constant.

The natural logarithm in the velocity function leads us to use exponentiation. Solving these integrations gives us the function \( v(t) = v_0 e^{-\frac{k}{m}t} \), representing the decline of velocity over time.
Further integration of \( v(t) \) allows us to determine the position function \( s(t) \), demonstrating how integration ties together time-dependent changes in both velocity and position in dynamic systems.

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