Problem 17 - Advanced Engineering Mathematics

Question

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 \(\mathrm{DE}\) 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) The solution is velocity in terms of time. (b) Terminal velocity is \( v_t = \sqrt{\frac{mg}{k}} \). (c) Distance is found by integrating velocity over time.

Step 1: Analyze the Differential Equation

The given differential equation is \( m\frac{dv}{dt} = mg - kv^2 \). This is a first-order nonlinear ordinary differential equation (ODE). The term \( mg \) represents the gravitational force, and \( kv^2 \) represents the air resistance, which is proportional to the square of the velocity.

Step 2: Separation of Variables

Rearrange the given differential equation to separate the variables. We get:\[m\frac{dv}{dt} = mg - kv^2 \]Dividing both sides by \( mg - kv^2 \) and multiplying by \( dt \), we can write:\[\frac{m}{mg - kv^2} dv = dt \]

Step 3: Integrate Both Sides

Integrate the separated equation:\[\int \frac{m}{mg - kv^2} dv = \int dt \]The integration can be done using substitution methods. Set \( u = \sqrt{k}v \), then \( du = \sqrt{k} dv \). The equation becomes:\[\frac{1}{\sqrt{k}} \int \frac{1}{g - u^2} du = t + C\]

Step 4: Solve the Integral

The integral \( \int \frac{1}{g - u^2} du \) is similar to a standard inverse hyperbolic function, yielding:\[\frac{1}{\sqrt{k}} \tanh^{-1}\left(\frac{u}{\sqrt{g}}\right) = t + C \]Convert back in terms of \( v \) and solve for \( v(t) \).

Step 5: Apply Initial Condition

Use the initial condition \( v(0) = v_0 \) to find the constant \( C \). Substitute \( v = v_0 \) and \( t = 0 \) into the integrated function to solve for \( C \).

Step 6: Determine Terminal Velocity

The terminal velocity is the velocity as \( t \to \infty \). At terminal velocity, the acceleration is zero, so:\[m g = k v_t^2\]Solve for \( v_t \):\[v_t = \sqrt{\frac{mg}{k}} \]

Step 7: Relationship Between Distance and Velocity

The relationship is given by \( \frac{ds}{dt} = v(t) \). Integrate the expression for \( v(t) \) from part (a) over time to find \( s(t) \). Use the initial condition \( s(0) = 0 \) to find the specific solution for \( s(t) \).

Key Concepts

terminal velocity

When an object falls through a fluid like air, it experiences air resistance that opposes its motion. As it falls faster, this resistance increases until it balances the force of gravity. Terminal velocity is reached when the gravity pulling downward is equal to the upward force of air resistance. At this point, the object ceases to accelerate and continues to fall at a constant speed.

This terminal velocity (\(v_t\)) is given by the equation \(v_t = \sqrt{\frac{mg}{k}}\), where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity, and \(k\) is the drag coefficient which quantifies air resistance.

Therefore, determining terminal velocity is crucial in studying the motion of objects through the air as it defines the maximum speed an object can reach.

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 proportional to some power of the velocity. In this particular problem, air resistance is proportional to the square of the velocity \(kv^2\).

• Greater speed causes greater air resistance.
• It depends on factors like the shape of the object, the density of air, and the object's speed.

Air resistance plays a significant role in determining the terminal velocity and affects how quickly an object accelerates when falling. As velocity increases, so does the force of air resistance, reducing the net force on the object until terminal velocity is achieved.

separation of variables

Separation of variables is a technique used to solve differential equations by separating the variables into distinct sides of the equation. This allows for easier integration and solving for one variable in terms of another.

For the differential equation given, \(m\frac{dv}{dt} = mg - kv^2\), we separate the variables by rearranging the terms to isolate \(dv\) on one side and \(dt\) on the other. This method transforms the equation into \(\frac{dv}{mg-kv^2} = \frac{dt}{m}\) which can then be integrated to find a solution in terms of time \(t\) and velocity \(v\).

Separation of variables is efficient because it breaks down a complex equation into simpler parts, making it easier to solve.

first-order nonlinear ODE

In mathematics, a first-order nonlinear ordinary differential equation (ODE) refers to an equation where the highest derivative is the first derivative, and the equation is not linear in terms of function or its derivatives. The differential equation from the problem, \(m\frac{dv}{dt} = mg - kv^2\), is an example of such an equation.

Characteristics of first-order nonlinear ODEs include:

• The equation involves products or powers of the function or its derivatives.
• They may not have explicit general solutions and often require numerical methods or special techniques like separation of variables for solution.

Understanding the properties and solutions of such differential equations is crucial in physics and engineering for modeling real-world phenomena.

initial condition

An initial condition in the context of differential equations is an extra condition that allows the determination of a specific solution from a set of possible solutions. In other words, it is a given value of the unknown function at a specific point, often provided to "anchor" the solution to the problem's physical context.

In the exercise, the initial condition is given as \(v(0)=v_0\), meaning the object's initial velocity is \(v_0\) at time \(t=0\). Applying this initial condition allows us to solve for the integration constant during the integration process, ensuring that our solution is not just general but tailored to match the specific context of the situation described.

Initial conditions are essential for solving differential equations with real-world applications because they ensure the solution accurately reflects the scenario.