Problem 65 - Calculus Early Transcendentals

Question

Suppose that a particle moving along the \(x\) -axis encounters a resisting force that results in an acceleration of \(a=d v / d t=-\frac{1}{32} v^{2}\) If \(x=0 \mathrm{cm}\) and \(v=128 \mathrm{cm} / \mathrm{s}\) at time \(t=0,\) find the velocity \(v\) and position \(x\) as a function of \(t\) for \(t \geq 0 .\)

Step-by-Step Solution

Answer

Velocity: \(v(t) = \frac{128}{4t + 1}\). Position: \(x(t) = 32 \ln(4t + 1)\).

Step 1: Relate Acceleration to Velocity

The given acceleration is expressed as \(a = \frac{dv}{dt} = -\frac{1}{32}v^2\). Our task is to find the velocity \(v\) as a function of time \(t\), which leads us to solve a differential equation. This involves separating variables to integrate \(\frac{dv}{v^2} = -\frac{1}{32}dt\).

Step 2: Integrate the Differential Equation

Integrate both sides: \(\int \frac{dv}{v^2} = -\int \frac{1}{32} dt\). This results in \(-\frac{1}{v} = -\frac{1}{32}t + C\), where \(C\) is the constant of integration.

Step 3: Solve for the Constant of Integration

Use the initial condition \(v(0) = 128\) to find \(C\). Substituting into the equation, \(-\frac{1}{128} = C\). Thus, our equation becomes \(-\frac{1}{v} = -\frac{1}{32}t - \frac{1}{128}\).

Step 4: Solve for Velocity as a Function of Time

Rearrange the modified equation \(-\frac{1}{v} = -\frac{1}{32}t - \frac{1}{128}\) to solve for \(v\). Multiply through by \(v\) to get \(v = \frac{1}{\frac{1}{32}t + \frac{1}{128}}\). Simplify the expression to obtain \(v(t) = \frac{128}{4t + 1}\).

Step 5: Relate Velocity to Position

Now express \(x\) as a function of time using the velocity equation. Recall \(v = \frac{dx}{dt}\), which implies \(dx = v\, dt\). Substituting the velocity function gives \(dx = \frac{128}{4t + 1}dt\).

Step 6: Integrate to Find Position as a Function of Time

Integrate the expression \(x(t) = \int \frac{128}{4t + 1}dt\). Use substitution to solve the integral, leading to \(x(t) = 32 \ln(|4t + 1|) + D\), where \(D\) is the constant of integration. Using the initial condition \(x(0) = 0\), solve for \(D\), leading to \(D = 0\).

Step 7: Write the Final Expression for Position

Finally, we have \(x(t) = 32 \ln(4t + 1)\). This gives us the position as a function of time.

Key Concepts

Integration

Integration is a fundamental mathematical concept that helps us accumulate values over intervals. It is crucial in solving differential equations, which often describe how a function changes, such as the velocity of a moving object. In our problem, the integration process is used twice:

• First, to go from the object's acceleration equation to find the velocity as a function of time.
• Second, to determine the position of the particle along the line from the velocity function.

To integrate the given differential equation, we started by separating variables, which involves rearranging the equation so that each side contains only one variable. This resulted in the equation \[ \int \frac{dv}{v^2} = -\int \frac{1}{32} dt. \]Integrating both sides helps us find an expression relating velocity \(v\) with the time \(t\). By solving this integral, we determine how the velocity decreases over time due to the resisting force.

Initial Conditions

Initial conditions are vital in finding specific solutions to differential equations. They provide the necessary information to calculate the constants that appear after integration. In our exercise, two initial conditions were given:

• When \(t = 0\), the velocity \(v\) was 128 cm/s.
• Additionally, at \(t = 0\), the position \(x\) was 0 cm.

Using the initial condition for velocity, we found the constant \(C\) after integrating the velocity equation. By substituting \(v(0) = 128\) into the equation \[ -\frac{1}{v} = -\frac{1}{32}t + C, \]we can solve for \(C\), which allowed us to write the velocity as a clear function of time. Similarly, the initial condition for position guided us in confirming that the constant of integration \(D\) for the position equation was zero. So, initial conditions enable us to translate general solutions from integration into practical, real-world contexts.

Motion Along a Line

Motion along a line describes how an object travels in a single dimension, usually along the x-axis in our mathematical models. This type of motion is often simplified into problems involving velocity and position functions of time, which are solutions of differential equations.
In our case, the object is experiencing a resisting force causing acceleration that is inversely proportional to the square of its velocity. This is summarized by the acceleration equation \[ a = \frac{dv}{dt} = -\frac{1}{32}v^2. \]By integrating this and using the initial conditions, we derived both a velocity function \(v(t) = \frac{128}{4t + 1}\) and a position function \(x(t) = 32 \ln(4t + 1)\). These functions illustrate the complete motion of the particle along the line as time progresses. Understanding these functions helps visualize how the particle's speed decreases over time due to the resisting force and how its position changes along the line.