Question

An object in free fall may be modeled by assuming that the only forces at work are the gravitational force and air resistance. By Newton's Second Law of Motion (mass \(\times\) acceleration \(=\) the sum of the external forces), the velocity of the object satisfies the differential equation $$\underbrace {m}_{\text {mass}}\quad \cdot \underbrace{v^{\prime}(t)}_{\text {acceleration }}=\underbrace {m g+f(v)}_{\text {external forces}}$$ where \(f\) is a function that models the air resistance (assuming the positive direction is downward). One common assumption (often used for motion in air) is that \(f(v)=-k v^{2},\) where \(k>0\) is a drag coefficient. a. Show that the equation can be written in the form \(v^{\prime}(t)=g-a v^{2},\) where \(a=k / m\) b. For what (positive) value of \(v\) is \(v^{\prime}(t)=0 ?\) (This equilibrium solution is called the terminal velocity.) c. Find the solution of this separable equation assuming \(v(0)=0\) and \(0 < v^{2} < g / a\) d. Graph the solution found in part (c) with \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) \(m=1,\) and \(k=0.1,\) and verify that the terminal velocity agrees with the value found in part (b).

Short answer

Question: Rewrite the given differential equation in the form \(v^{\prime}(t)=g-a v^{2}\), determine the positive equilibrium velocity, and solve the separable equation for the provided initial condition. Verify your findings by graphing the solution and discussing the terminal velocity. Answer: The given differential equation can be rewritten as \(v^{\prime}(t)=g-a v^{2}\). The positive equilibrium velocity is calculated as \(v = \sqrt{\frac{g}{a}}\). The separable equation can then be solved, resulting in the function: \(v(t)=\frac{1}{\sqrt{a}}\tan(\sqrt{a}gt)\). By graphing this solution, the terminal velocity can be observed to agree with the calculated value from part (b), as the graph approaches the terminal velocity \(\sqrt{98}\) as the object falls.

Step-by-step solution

Rewriting the Equation

Given the differential equation: \[m v^{\prime}(t) = mg + f(v)\] Where \(f(v) = -kv^2\) and \(a = \frac{k}{m}\). We can rewrite the equation as follows: \[v^{\prime}(t) = \frac{mg - kv^2}{m}\] and since \(a=k / m\), we have: \[v^{\prime}(t) = g - av^2\] #b. Determine the positive equilibrium velocity.#

Equilibrium Velocity

At equilibrium, the object's acceleration (\(v^{\prime}(t)\)) is 0. Therefore, we can set the rewritten equation to 0 and solve for \(v\): \[0 = g - av^2\] Divide both sides by \(a\): \[0 = \frac{g}{a} - v^2\] Rearrange: \[v^2 = \frac{g}{a}\] Take the square root to find the positive equilibrium velocity (\(v\)): \[v = \sqrt{\frac{g}{a}}\text{ (positive value)}\] #c. Solve the separable equation assuming an initial condition.#

Solving the Separable Equation

We have the equation: \[v^{\prime}(t) = g - av^2\] Let's rewrite it in a separable form: \[\int\frac{1}{g - av^2}dv = \int dt\] Integrating both sides and assuming \(v(0)=0\) and \(0<v^2<\frac{g}{a}\), we have \[\frac{1}{\sqrt{a}} \arctan(\sqrt{a}v) = gt + C\] To find \(C\), set \(t=0\) and \(v(0)=0\): \[\frac{1}{\sqrt{a}} \arctan(\sqrt{a}(0)) = g(0) + C\] So, \(C=0\). Thus, \[v(t)=\frac{1}{\sqrt{a}}\tan(\sqrt{a}gt)\] #d. Graph the solution and verify that the terminal velocity agrees with the value found in part (b).#

Terminal Velocity and Graph Solution

Using the provided values \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\), \(m=1\), and \(k=0.1\), let's calculate the terminal velocity from part (b): \[v = \sqrt{\frac{g}{a}} = \sqrt{\frac{9.8}{\frac{0.1}{1}}} = \sqrt{98}\] Now, let's graph the solution using the formula we obtained in part (c): \[v(t)=\frac{1}{\sqrt{a}}\tan(\sqrt{a}gt)=\frac{1}{\sqrt{0.1}}\tan(\sqrt{0.1}\cdot 9.8t)\] As the object falls, the graph of \(v(t)\) will approach the terminal velocity \(\sqrt{98}\) we calculated in part (b). By observing the graph, it can be seen that the terminal velocity agrees with the calculated value, confirming the result from part (b). (Note: Graphing is not supported here; please use graphing tools like Desmos, GeoGebra, or even a graphing calculator to visualize the function.)

Key concepts

Newton's Second Law

Newton's Second Law of Motion states that the force acting on an object is equal to the mass of that object times its acceleration. Mathematically, this is expressed as \( F = ma \). In the context of the given exercise, this principle is used to model the motion of an object in free fall. The external forces considered here are gravity and air resistance.

• Gravity is a constant force that pulls the object downward.
• Air resistance, on the other hand, acts in the opposite direction of the velocity.

By balancing these forces, we can derive a differential equation that describes the behavior of the object's velocity over time. This forms the foundation for solving many real-world problems involving motion.

Air Resistance Modeling

In physics, air resistance is an important factor when modeling the motion of objects through the air. It is a type of friction that opposes the motion of an object. In the exercise, air resistance is modeled using a function \( f(v) = -kv^2 \).

• The negative sign indicates that the force opposes the direction of motion.
• The parameter \( k \) is the drag coefficient, which depends on several factors like the shape and size of the object.
• The square of the velocity \( v^2 \) implies that air resistance increases with the square of the speed.

This quadratic relationship is typical for objects moving at high speeds, where drag depends heavily on velocity. This makes the model more complex but realistic, reflecting how real-life situations operate.

Terminal Velocity

Terminal velocity is a key concept in the study of motion under forces such as gravity and air resistance. It is the constant speed that an object reaches when the force of gravity is balanced by the drag force due to air resistance, resulting in zero acceleration.
When the net external force is zero, the object's velocity becomes constant at this terminal speed. In the given exercise, the terminal velocity can be found by setting the acceleration to zero and solving the equation: \[ 0 = g - av^2 \]. This gives the terminal velocity as \( v = \sqrt{\frac{g}{a}} \). This signifies that the object no longer speeds up and continues to fall at a steady speed. Terminal velocity is vital in understanding many phenomena, such as how parachutes work and why raindrops don’t cause damage when they hit the ground.

Separable Equations

Separable equations are a class of differential equations that can be solved algebraically by separating the variables before integrating. In the case of motion with air resistance, the differential equation is given as \( v'(t) = g - av^2 \). This is rearranged into a separable form:

• Separate variables: \( \int \frac{1}{g - av^2} dv = \int dt \)
• Integrate both sides: The left side concerning \( v \) and the right side concerning \( t \).

By integrating, we can find the general solution that represents the velocity of the object as a function of time. This technique simplifies the complex relationship between force, mass, and acceleration into more manageable algebraic expressions that can be easily solved.

Free Fall Motion

Free fall motion describes the movement of objects solely under the influence of gravity. In the absence of air resistance, objects in free fall would accelerate indefinitely under gravity. However, when air resistance is present, as modeled in this exercise, the velocity reaches an equilibrium point or terminal velocity where the net force becomes zero.
The exercise illustrates how to compute the velocity of an object from rest in free fall with air resistance by solving a specific type of differential equation.

• With no other forces acting besides gravity and air resistance, the object quickly reaches terminal velocity.
• This illustrates how environmental factors like air affect motion, making simple forces more complex in real-life applications.

Understanding free fall with air resistance is crucial for predicting the behavior of falling objects in practical situations like skydiving or atmospheric re-entry of spacecraft.