Problem 66 - Calculus Early Transcendentals

Question

Suppose that a particle moving along the \(x\) -axis encounters a resisting force that results in an acceleration of \(a=d v / d t=-0.02 \sqrt{v}\). Given that \(x=0 \mathrm{cm}\) and \(v=9\) \(\mathrm{cm} / \mathrm{s}\) 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: \( v = (3 - 0.01t)^2 \), Position: \( x = 9t - 0.03t^2 + 0.0000333t^3 \).

Step 1: Rearrange the acceleration equation

The acceleration given is \( a = \frac{dv}{dt} = -0.02 \sqrt{v} \). Rearrange this to express the derivative of velocity with respect to velocity and time: \( dv = -0.02 \sqrt{v} \, dt \).

Step 2: Separate and integrate the velocity equation

Separate the variables to isolate terms involving \(v\) on one side: \( \frac{dv}{\sqrt{v}} = -0.02 \, dt \). Integrate both sides to find \( v(t) \). The integral becomes: \( \int \frac{dv}{\sqrt{v}} = \int -0.02 \, dt \).

Step 3: Solve the integrals

Perform the integration: the left side becomes \( 2\sqrt{v} \) and the right side becomes \( -0.02t + C \), where \( C \) is the integration constant. Thus, \( 2\sqrt{v} = -0.02t + C \).

Step 4: Apply initial conditions

Use the initial condition \( v(0) = 9 \) to determine \( C \). When \( t = 0 \), \( 2\sqrt{9} = C \), so \( C = 6 \). Thus, \( 2\sqrt{v} = 6 - 0.02t \).

Step 5: Solve for velocity function

Rearrange to solve for \( v \): \( \sqrt{v} = 3 - 0.01t \), so \( v = (3 - 0.01t)^2 \) for \( t \geq 0 \).

Step 6: Set up the differential equation for position

We know \( v = \frac{dx}{dt} = (3 - 0.01t)^2 \). Rearrange to express \( dx \) in terms of \( t \): \( dx = (3 - 0.01t)^2 \, dt \).

Step 7: Integrate to find the position function

Integrate both sides to find \( x(t) \): \( \int dx = \int (3 - 0.01t)^2 \, dt \). Perform this integration using the expansion \( (3 - 0.01t)^2 = 9 - 0.06t + 0.0001t^2 \).

Step 8: Solve the position integral

Integrate term by term: \( x(t) = 9t - 0.03t^2 + 0.0000333t^3 + C \).

Step 9: Apply initial conditions for position

Apply the initial condition \( x(0) = 0 \) to solve for the constant \( C \). Since all terms vanish for \( t = 0 \), we find \( C = 0 \). Thus, \( x(t) = 9t - 0.03t^2 + 0.0000333t^3 \).

Key Concepts

Particle Motion

Particle motion refers to the behavior and path followed by an object as it moves through space. In the case of differential equations, particle motion often involves understanding how variables like velocity and position change over time.

• Velocity: The rate of change of position with respect to time.
• Acceleration: The rate of change of velocity with respect to time.
• Position: The location of the particle at any given time.

In our example, the particle is subjected to a resisting force that results in a unique movement behavior along the x-axis. The motion is characterized by a specific initial velocity and acceleration determined by the resisting force. Understanding these dynamics is key to predicting the particle's future position and velocity.

Initial Value Problem

An initial value problem involves finding a function that satisfies a differential equation and also meets a specific initial condition. This condition provides a starting point for the solution. For any initial value problem, it is essential to know:

• Single or multiple initial conditions: These are the known values of variables at a specific point.
• Purpose: To provide a unique solution to the differential equation by eliminating any arbitrary constants.

In the context of our exercise, we are given that at time \( t = 0 \), the particle has a position \( x = 0 \) and a velocity \( v = 9 \) cm/s. This information enables us to solve the differential equations for both velocity and position, ensuring the results fit the conditions stated.

Separable Equations

Separable equations are a class of differential equations that can be split into two parts, each involving only one variable. The technique involves rearranging the equation so that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the opposite side. Once separated, each side can be integrated independently.
Here are the basic steps to solve separable equations:

• Rearrange the equation to isolate the variables on separate sides.
• Integrate both sides with respect to their respective variables.
• Apply any initial conditions to solve for the integration constant.

In the original solution, the velocity equation is made separable by isolating the velocity terms. This enables one to integrate each part and solve the equation under the given initial conditions. This approach provides a methodical means to solve for the particle's velocity and demonstrates the power of separation in differential equations.