Question

Free fall An object in free fall may be modeled by assuming the only forces at work are the gravitational force and air resistance. By Newton's Second Law of Motion (mass \cdot acceleration \(=\) the sum of external forces), the velocity of the object satisfies the differential equation $$\underbrace{m}_{\text {mass }} \cdot \underbrace{v^{\prime}(t)}_{\text {acceleration }}=\underbrace{m g+f(v)}_{\text {extermal forces }}$$ 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 \(f(v)=-k v^{2},\) for \(t \geq 0,\) 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 \(a=\frac{k}{m}\) 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

Short answer

Based on the provided information, the step-by-step solution to the problem has been detailed, and we have found that the terminal velocity of the object is approximately \(v_{\text{terminal}}=\sqrt{98}\). The graph of the velocity function \(v(t)\) with the given constants should become horizontal around this value, verifying our result.

Step-by-step solution

Rewrite the equation in the form \(v^{\prime}(t)=g-av^2\)

The given equation is $$ m\cdot v^{\prime}(t)=mg-k v^{2} $$ Now divide the equation by \(m\): $$ mv^{\prime}(t)=mg-kv^2 \Rightarrow v^{\prime}(t)=g-\frac{k}{m}v^{2} $$ Then, let \(a=\frac{k}{m}\). So, the equation becomes: $$ v^{\prime}(t)=g-av^2 $$

Determine the terminal velocity

We are asked to find the positive value of \(v\) when \(v^{\prime}(t)=0\). Using the equation from step 1: $$ 0=g-av^{2} $$ Rearrange the equation: $$ av^{2}=g \Rightarrow v^{2}=\frac{g}{a} $$ Take the square root of both sides: $$ v = \sqrt{\frac{g}{a}} $$ This is the terminal velocity.

Solve the differential equation using separation of variables

Now we need to solve the separable differential equation: $$ v^{\prime}(t)=g-av^{2} $$ Write this equation in terms of differentials: $$ \frac{dv}{dt}=g-av^{2} $$ Separate the variables: $$ \frac{dv}{g-av^{2}}=dt $$ Now integrate both sides: $$ \int \frac{dv}{g-av^{2}} = \int dt $$ Let's perform the integration on both sides. On the left side we need to perform a substitution. Let \(u=g-av^{2}\), then \(du=-2avdv\): $$ \int \frac{1}{-2a}\frac{du}{u} = \int dt $$ Now integrate with respect to the new variable, u, as well as t: $$ \frac{-1}{2a}\ln|u| = t+C $$ Now substitute \(u=g-av^{2}\) back into the equation: $$ \frac{-1}{2a}\ln|g-av^{2}| = t+C $$

Graph the solution end verify the terminal velocity

To find the function \(v(t)\), solve the equation from the previous step for \(v\): $$ \frac{-1}{2a}\ln|g-av^{2}| = t+C \Rightarrow \ln|g-av^{2}| = -2a(t+C) \Rightarrow |g-av^{2}|=e^{-2a(t+C)} $$ Let \(D=e^{-2aC}\): $$ g-av^{2}=D e^{-2at} \Rightarrow av^{2}=g-De^{-2at} \Rightarrow v^{2}=\frac{g}{a}-D'e^{-2at}, $$ where \(D'=\frac{D}{a}\). Now take the square root: $$ v(t) = \sqrt{ \frac{g}{a}-D'e^{-2at}} $$ This is our function for velocity. To find the terminal velocity from the graph, we need the graph to become horizontal (i.e., the velocity doesn't change). This means, \(v^{\prime }(t)=0\). Using the previous results, we know that the terminal velocity is \(v=\sqrt{\frac{g}{a}}\). With the given parameters \(g=9.8\,\text{m/s}^{2}\), \(m=1\), and \(k=0.1\), we can find the terminal velocity: $$ v_{\text{terminal}} = \sqrt{ \frac{9.8}{0.1}} \approx \sqrt{98} $$ Plot the function \(v(t)\) with the given constants and verify that the graph becomes horizontal around this value.

Key concepts

Newton's Second Law

Newton's Second Law of Motion provides a fundamental principle in understanding how objects move. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, this can be expressed as: \[ F = m imes a \] Where \( F \) is the net external force, \( m \) is the mass, and \( a \) is the acceleration of the object. This law is crucial when analyzing the motion of objects under the influence of various forces, such as gravity and air resistance.
In the context of a freely falling object, the forces involved are gravity and air resistance. Gravity pulls the object downward, while air resistance acts in the opposite direction. The net force, therefore, affects the object's acceleration, altering its velocity over time. In our problem, this results in the differential equation: \[ m imes v'(t) = mg + f(v) \] where \( v'(t) \) denotes the acceleration. This equation serves as the foundation for solving more complex motion-related problems.

Terminal Velocity

Terminal velocity is a key concept in physics, especially when discussing objects in free fall. It refers to the constant speed that an object reaches when the forces of gravity and air resistance are balanced. At terminal velocity, acceleration ceases, and the velocity remains stable.
In mathematical terms, this occurs when the change in velocity over time is zero, \( v'(t) = 0 \). For our problem, setting the differential equation to zero gives: \[ g - av^2 = 0 \] Solving this gives the terminal velocity: \[ v = \sqrt{\frac{g}{a}} \] Terminal velocity depends on the drag coefficient \( k \), mass \( m \), and gravitational acceleration \( g \). A smaller drag or a larger mass results in a higher terminal velocity. This concept is essential in disciplines such as parachuting, where reaching a stable, safe speed is crucial.

Air Resistance

Air resistance, or drag, plays a significant role in the dynamics of any object moving through a fluid, such as air. It acts against the direction of motion and generally depends on factors like the object's velocity, shape, and surface area.
The equation for air resistance commonly used in problems involving freefall is:- Linear drag: \( f(v) = -kv \)- Quadratic drag: \( f(v) = -kv^2 \) In the quadratic form, which is applicable to objects falling at high speeds, the drag force increases with the square of velocity, making it a crucial force to consider.
In this exercise, we assume quadratic air resistance, characterized by the equation \[ f(v) = -kv^2 \] The negative sign indicates that the force acts in a direction opposite to motion. Understanding air resistance is essential for predicting how objects will move through any medium where drag is a factor.

Separable Equations

Separable equations are a type of differential equation that can be rearranged into a form allowing us to separate variables on each side. This makes them relatively simpler to integrate and solve. The basic form of a separable equation is: \[ \frac{dy}{dx} = g(x)h(y) \] Rearranging gives: \[ \frac{1}{h(y)} dy = g(x) dx \] This allows both sides to be integrated separately.
In our exercise, the equation \[ v'(t) = g-av^2 \] can be written as \[ \frac{dv}{dt} = g - av^2 \] By rearranging, we can separate the variables: \[ \frac{dv}{g-av^2} = dt \] Integrating both sides gives a solution for velocity \( v(t) \). This method is incredibly useful in physics and engineering for splitting complex problems into more manageable parts that can be solved step by step.