Problem 17 - Advanced Engineering Mathematics

Question

Air Resistance A differential equation governing the velocity \(v\) of a falling mass \(m\) subjected to air resistance proportional to the square of the instantaneous velocity is $$ m \frac{d v}{d t}=m g-k v^{2} $$ where \(k>0\) is the drag coefficient. The positive direction is downward. (a) Solve this equation subject to the initial condition \(v(0)=v_{0}\). (b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the DE in Problem 39 in Exercises 2.1. (c) If distance \(s\), measured from the point where the mass was released above ground, is related to velocity \(v\) by \(d s / d t=v(t)\), find an explicit expression for \(s(t)\) if \(s(0)=0\).

Step-by-Step Solution

Answer

(a) Use separation of variables to solve for \( v(t) \). (b) Terminal velocity is \( v_t = \sqrt{\frac{m g}{k}} \). (c) Integrate \( v(t) \) to find \( s(t) \).

Step 1: Write the Differential Equation in Terms of v

We start with the differential equation given by \[ m \frac{d v}{d t} = m g - k v^2. \] This can be rewritten as \[ \frac{d v}{d t} = g - \frac{k}{m} v^2. \] This equation can be integrated to find the function \( v(t) \).

Step 2: Separate Variables

Separate variables in the differential equation to facilitate integration. We can write \[ \frac{d v}{g - \frac{k}{m} v^2} = d t. \] This aligns the terms with respect to \( v \) on one side and \( t \) on the other side.

Step 3: Integrate Both Sides

Integrate both sides. For the left-hand side, we use partial fraction decomposition if necessary. The result gives \[ \int \frac{1}{g - \frac{k}{m} v^2} \, dv = \int d t. \] Solving this and applying the initial condition \( v(0) = v_0 \), we can find an expression for \( v(t) \).

Step 4: Solve the Integration

Integrate and solve using a trigonometric substitution like \( v = \sqrt{\frac{m g}{k}} \sin(\theta) \) or recognize the form and directly integrate. We find \[ t + C_1 = \frac{m}{2 \sqrt{g k}} \ln \left| \frac{\sqrt{g k} + \sqrt{k v_0^2 - m g}}{\sqrt{g k} - \sqrt{k v_0^2 - m g}} \right|. \] Applying the initial condition helps solve for the constant \( C_1 \).

Step 5: Find the Terminal Velocity

For terminal velocity, the acceleration should be 0. Solving \( m g - k v^2 = 0 \), we derive \( v_t = \sqrt{\frac{m g}{k}} \). Thus, as \( t \to \infty \), \( v(t) \rightarrow \sqrt{\frac{m g}{k}} \).

Step 6: Determine Position Function s(t)

Given \( \frac{ds}{dt} = v(t) \), integrate \( v(t) \) with respect to \( t \) once it's determined, applying the initial condition \( s(0) = 0 \). This yields \( s(t) = \int_0^t v(u) \, du \), resulting in the explicit expression for \( s(t) \).

Key Concepts

Air Resistance

Air resistance, also known as drag, is a force that acts opposite to the relative motion of an object moving through the air. It is critical to understanding how objects fall through the atmosphere.
It increases with the square of the object's velocity, making it a quadratic force. In our differential equation, it is represented by the term \( - k v^2 \).
This means as the velocity increases, the force due to air resistance increases, eventually balancing the gravitational force. This balance affects the object's motion, particularly when approaching terminal velocity.

Terminal Velocity

Terminal velocity refers to the maximum speed that an object can achieve as it falls through a fluid, like air. It's the point where the force of gravity is balanced by the force of drag.
In mathematical terms, it is found by setting the net force to zero, resulting in no acceleration, hence:

• \( m g = k v_t^2 \)
• Solving for \( v_t \) gives \( v_t = \sqrt{\frac{m g}{k}} \).

This velocity is constant and is important because no matter how much longer the body falls, its speed doesn't increase further if undisturbed.

Initial Conditions

When solving differential equations, initial conditions are essential. They provide specific values at the beginning (often \( t = 0 \)) which help to determine exact solutions.
In our problem, the initial condition is given by \( v(0) = v_0 \).

• These ensure the integration constants can be defined.
• They make the solution unique and applicable to real-world scenarios.

Without applying these conditions, our solution could be deemed incomplete or overly general.

Separation of Variables

Separation of variables is a method for solving differential equations by separating the variables, typically placing all terms involving one variable on one side of the equation.
In the exercise, we rearrange the differential equation to isolate \( v \) terms with \( dv \) on one side and \( t \) with \( dt \) on the other:
\[ \frac{d v}{g - \frac{k}{m} v^2} = d t \]
This allows for straightforward integration across both sides, making it simpler to solve complex equations step-by-step. It's a widely used technique as it is easy to manage once variables are separated and simplifies solving intricate equations.