Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

The differential equation for a falling body near the earth's surface with air resistance proportional to the velocity \(v\) is \(d v / d t=-g-a v\), where \(g=32\) feet per second per second is the acceleration of gravity and \(a>0\) is the drag coefficient. Show each of the following: (a) \(v(t)=\left(v_{0}-v_{\infty}\right) e^{-a t}+v_{\infty}\), where \(v_{0}=v(0)\), and $$ v_{\infty}=-g / a=\lim _{t \rightarrow \infty} v(t) $$ is the so-called terminal velocity. (b) If \(y(t)\) denotes the altitude, then $$ y(t)=y_{0}+t v_{\infty}+(1 / a)\left(v_{0}-v_{\infty}\right)\left(1-e^{-a t}\right) $$

Step-by-Step Solution

Answer

For (a) use an integrating factor to solve for velocity. For (b), integrate velocity to get altitude.

Step 1: Understanding the Differential Equation

We start with the differential equation given for velocity: \( \frac{dv}{dt} = -g - av \). Here, \(g\) is the acceleration due to gravity (32 \( \frac{ft}{s^2} \)), and \(a\) is the positive drag coefficient.

Step 2: Solve the Differential Equation

To solve \( \frac{dv}{dt} = -g - av \), we rearrange it to standard form: \( \frac{dv}{dt} + av = -g \).This can be solved using an integrating factor. The integrating factor \( I(t) \) is \( e^{at} \). Multiply through by the integrating factor:\[ e^{at} \frac{dv}{dt} + ae^{at}v = -ge^{at}. \] The left side is the derivative of \( e^{at} \cdot v \), so we integrate:\[ \int \frac{d}{dt}(e^{at}v) \, dt = \int -ge^{at} \, dt. \]The left side integrates to \( e^{at}v \), and the right side to \( -\frac{g}{a} e^{at} + C \). Thus,\[ e^{at}v = -\frac{g}{a}e^{at} + C. \] Solving for \( v \), we get:\[ v(t) = -\frac{g}{a} + Ce^{-at}. \]

Step 3: Determine the Value of the Constant C

We apply the initial condition \( v(0) = v_0 \) to find \( C \). At \( t = 0 \),\[ v(0) = -\frac{g}{a} + C = v_0 \]Solving for \( C \), we get \( C = v_0 + \frac{g}{a} \). Substitute back to get:\[ v(t) = -\frac{g}{a} + \left(v_0 + \frac{g}{a}\right)e^{-at}. \]

Step 4: Express Terminal Velocity

Simplify the expression:\[ v(t) = v_0 e^{-at} + \left( -\frac{g}{a} (1 - e^{-at}) \right), \]which simplifies to:\[ v(t) = (v_0 - v_{\infty}) e^{-at} + v_{\infty}, \]where \( v_{\infty} = -\frac{g}{a} \).

Step 5: Express Altitude Equation

Start with \( \, \frac{dy}{dt} = v(t) \), where \( y(t) \) is the altitude. Substituting in our velocity expression,\[ \frac{dy}{dt} = (v_0 - v_{\infty}) e^{-at} + v_{\infty}. \]Integrate both sides with respect to \( t \):\[ y(t) = \int \left( (v_0 - v_{\infty})e^{-at} + v_{\infty} \right) dt \]The integral yields:\[ y(t) = (v_0 - v_{\infty}) \left( -\frac{1}{a} e^{-at} \right) + v_{\infty}t + C_2. \]

Step 6: Apply Initial Altitude Condition

Apply the initial condition \( y(0) = y_0 \) to find \( C_2 \).At \( t = 0 \):\[ y_0 = -\frac{1}{a}(v_0 - v_{\infty}) + C_2 .\]Solving for \(C_2\) gives:\[ C_2 = y_0 + \frac{1}{a}(v_0 - v_{\infty}). \]Substitute back to get: \[ y(t) = y_0 + v_{\infty}t + \frac{1}{a}(v_0 - v_{\infty})(1 - e^{-at}). \]

Key Concepts

Velocity Equations

Velocity equations help describe the motion of objects and are crucial when discussing physics problems, such as a body falling near the earth's surface. These equations take into account various forces acting on an object. Here, we focus on the velocity equation derived from a differential equation representing a falling body with air resistance.

The differential equation is given as \(\frac{dv}{dt} = -g - av\), where \(g\) is the acceleration due to gravity, and \(a\) is the drag coefficient. This equation indicates that the rate of change of velocity over time depends on gravitational force and air resistance.

• Gravity contributes a constant downward acceleration of 32 feet per second squared.
• The air resistance, proportional to the velocity \(v\), adds a deceleration factor, reducing the object's speed as it increases.

Understanding how these factors interact helps in predicting the motion of the object accurately and is an essential aspect of calculating velocity in physics.

Terminal Velocity

Terminal velocity is a key concept in physics, referring to the highest speed an object reaches as it falls. In our scenario, the object achieves terminal velocity when the gravitational force pulling it downward is balanced equally by the air resistance pushing it upward. When these forces equalize, the velocity becomes constant, and further acceleration ceases.

In the differential equation: \(dv/dt = -g - av\), terminal velocity \(v_{\infty}\) occurs when \(dv/dt = 0\). Solving gives us \(v_{\infty} = -\frac{g}{a}\). This formula shows that terminal velocity depends inversely on the drag coefficient.

• Higher drag coefficients (\(a\)) lead to lower terminal velocities.
• Larger gravitational forces result in higher terminal velocities.

Terminal velocity is crucial in various fields, including parachuting, where knowing the highest speed during free fall can be vital for safety and design implications.

Altitude in Physics

In physics, altitude refers to an object's height above a reference level, often the earth's surface. It is a critical parameter when studying the motion of objects like projectiles or falling bodies.

The altitude equation derived in our problem is: \(y(t) = y_0 + v_{\infty}t + \frac{1}{a}(v_0 - v_{\infty})(1 - e^{-at})\). This equation considers initial conditions and the effects of drag force over time.

• The initial altitude, \(y_0\), is where the object starts its fall.
• \(v_{\infty}t\) indicates linear growth with time due to terminal velocity when drag force balances with gravity.
• The term \(\frac{1}{a}(v_0 - v_{\infty})(1 - e^{-at})\) accounts for the impact of drag on height gain or loss before reaching terminal velocity.

Understanding altitude changes helps in applications like calculating landing points in aerodynamics, aiming for precise altitude adjustments in aerospace, and more.

Solving Differential Equations

Solving differential equations is fundamental in physics to predict systems' behavior involving rates of change. The exercise involves solving the equation \(\frac{dv}{dt} = -g - av\) to determine velocity over time for a falling body.

To solve such equations, integrate factors is a common method, as seen here:

• Rearrange into standard form: \(\frac{dv}{dt} + av = -g\).
• Use the integrating factor \(e^{at}\) to simplify integration by turning it into a derivative of a product.
• Integrate both sides and solve for \(v(t)\).

Applying initial conditions helps finalize the solution by determining integration constants, rendering the equation meaningful in real-life scenarios. Mastery in solving such equations facilitates deeper understanding in fields like engineering, where many applications require solving complex differential equations to model real-world situations.