Problem 15 - Thomas Calculus

Question

Skydiving If a body of mass \(m\) falling from rest under the action of gravity encounters an air resistance proportional to the square of velocity, then the body's velocity \(t\) seconds into the fall satisfies the equation $$m \frac{d v}{d t}=m g-k v^{2}, \quad k > 0$$ where \(k\) is a constant that depends on the body's aerodynamic properties and the density of the air. (We assume that the fall is too short to be affected by changes in the air's density.) \begin{equation}\begin{array}{l}{\text { a. Draw a phase line for the equation. }} \\ {\text { b. Sketch a typical velocity curve. }} \\ {\text { c. For a } 110 \text { -lb skydiver }(m g=110) \text { and with time in seconds }} \\ {\text { and distance in feet, a typical value of } k \text { is } 0.005 . \text { What is the }} \\ {\text { diver's terminal velocity? Repeat for a } 200-\text { -lb skydiver. }}\end{array}\end{equation}

Step-by-Step Solution

Answer

The terminal velocity is approximately 148.32 ft/s for a 110-lb skydiver and 200 ft/s for a 200-lb skydiver.

Step 1: Identify Equilibrium Points

The equilibrium points occur when the derivative \( \frac{dv}{dt} \) is zero. Setting \( m g - k v^2 = 0 \), we find that the equilibrium velocity \( v_e \) satisfies: \[m g = k v_e^2 \]Solving for \( v_e \), we get \( v_e = \pm \sqrt{\frac{m g}{k}} \), but since velocity is positive, \( v_e = \sqrt{\frac{m g}{k}} \).

Step 2: Draw the Phase Line

On a number line (phase line), we plot the equilibrium point \( v_e \). To the left of \( v_e \), the derivative \( \frac{dv}{dt} = m g - k v^2 > 0 \), meaning velocity is increasing. To the right of \( v_e \), \( \frac{dv}{dt} < 0 \), so velocity is decreasing. This shows that \( v_e \) is a stable equilibrium.

Step 3: Sketch Typical Velocity Curve

On a time vs. velocity graph, the curve starts from zero velocity and asymptotically approaches \( v_e \). The graph is an increasing curve that plateaus at \( v = v_e \), indicating that the velocity stabilizes at terminal velocity as time goes to infinity.

Step 4: Calculate Terminal Velocity for 110-lb Skydiver

For a 110-lb skydiver, \( mg = 110 \) and \( k = 0.005 \). The terminal velocity \( v_t \) is given by: \[v_t = \sqrt{\frac{110}{0.005}} = \sqrt{22000} \approx 148.32 \text{ ft/s}\]

Step 5: Calculate Terminal Velocity for 200-lb Skydiver

For a 200-lb skydiver, \( mg = 200 \) and \( k = 0.005 \). Using the same formula for terminal velocity, we have: \[v_t = \sqrt{\frac{200}{0.005}} = \sqrt{40000} = 200 \text{ ft/s}\]

Key Concepts

Air Resistance

When objects fall through the atmosphere, they experience a force known as air resistance or drag. This force acts opposite to the object's motion and is dependent on several factors including the velocity of the object, the density of the air, and the shape of the object. In our specific problem, air resistance is modeled as being proportional to the square of the velocity, which suggests a quadratic resistance.

This means that as the object's velocity increases, the air resistance increases quadratically. Mathematically, this is described by the equation \( F_{ ext{drag}} = -k v^2 \), where \( k \) is a positive constant that relates to the object's aerodynamic properties and air density.

Understanding air resistance is crucial because it not only affects how quickly an object reaches the ground but also defines the concept of terminal velocity, a state where the force of air resistance equals the force of gravity.

Velocity of Falling Objects

Velocity is the speed of something in a given direction. For objects in free fall, velocity increases as they fall under the influence of gravity. However, the presence of air resistance complicates this motion. Initially, when the object starts to fall, air resistance is relatively low, and the object accelerates due to gravity.

As the velocity increases, air resistance grows until it becomes significant enough to counterbalance the gravitational pull, causing the acceleration to cease. This is depicted by a curve that rises sharply at first and gradually approaches a plateau.

The mathematics of velocity changes are described by the differential equation \( m \frac{d v}{d t} = m g - k v^2 \). This represents the balance of forces acting on the object, where \( m g \) is the force due to gravity and \( k v^2 \) is the air resistance. Understanding this balance is key to predicting the velocity at any point in time during the object's fall.

Terminal Velocity

Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium (air, in this case) prevents further acceleration. This occurs when the force of gravity is equal and opposite to the force of air resistance.

In formula terms, terminal velocity \( v_t \) can be calculated when the net force is zero: \( m g = k v_t^2 \). Solving for \( v_t \), we find \( v_t = \sqrt{\frac{m g}{k}} \).

For example, a 110-lb skydiver with a drag coefficient \( k = 0.005 \) will have a terminal velocity of approximately 148.32 feet per second, while a 200-lb skydiver will achieve a terminal velocity of exactly 200 feet per second. Terminal velocity is influenced by factors such as mass, the drag coefficient, and the object's surface area. Comprehending terminal velocity is not only vital for understanding the dynamics of freefall but also in applications such as designing parachutes.