Problem 7 - Active Calculus

Question

When a skydiver jumps from a plane, gravity causes her downward velocity to increase at the rate of \(g \approx 9.8\) meters per second squared. At the same time, wind resistance causes her velocity to decrease at a rate proportional to the velocity. a. Using \(k\) to represent the constant of proportionality, write a differential equation that describes the rate of change of the skydiver's velocity. b. Find any equilibrium solutions and decide whether they are stable or unstable. Your result should depend on \(k\). c. Suppose that the initial velocity is zero. Find the velocity \(v(t)\). d. A typical terminal velocity for a skydiver falling face down is 54 meters per second. What is the value of \(k\) for this skydiver? e. How long does it take to reach \(50 \%\) of the terminal velocity?

Step-by-Step Solution

Answer

a. \( \frac{dv}{dt} = g - kv \). b. \( v = \frac{g}{k} \) stable. c. \( v(t) = \frac{g}{k} (1-e^{-kt}) \). d. \( k \approx 0.1815 \text{ s}^{-1} \). e. \( t \approx 3.82 \text{ s} \).

Step 1: Define the differential equation

Identify the forces acting on the skydiver: gravity increases velocity at a rate of 9.8 m/s\textsuperscript{2} and wind resistance decreases velocity at a rate proportional to the velocity, represented by -kv. Therefore, the differential equation combining these two forces is: \( \frac{dv}{dt} = g - kv \)

Step 2: Find equilibrium solutions

Set the right hand side of the differential equation to zero to find the equilibrium solutions: \( g - kv = 0 \). Solving for \( v \), the equilibrium solution is: \( v = \frac{g}{k} \). To determine stability, consider the sign of the derivative. If the velocity is slightly higher or lower than \( \frac{g}{k} \), the wind resistance term will adjust to bring it back, indicating that the equilibrium solution \( v = \frac{g}{k} \) is stable.

Step 3: Solve the differential equation with initial velocity 0

Rewrite the differential equation and solve it using separation of variables: \( \frac{dv}{dt} + kv = g \). Integrate both sides: \( v(t) = \frac{g}{k} + (C)e^{-kt} \). Apply the initial condition \( v(0) = 0 \) to find C: \( 0 = \frac{g}{k} + C \). Therefore, \( C = -\frac{g}{k} \). The solution is: \( v(t) = \frac{g}{k} (1 - e^{-kt}) \).

Step 4: Find the value of k for a terminal velocity of 54 m/s

Set the terminal velocity \( v = 54 \) m/s and solve for \( k \) using the equilibrium solution \( v = \frac{g}{k} \): \( 54 = \frac{9.8}{k} \). Solving for \( k \), we find: \( k = \frac{9.8}{54} \approx 0.1815 \text{ s}^{-1} \).

Step 5: Find the time to reach 50% of terminal velocity

Set the velocity \( v(t) = 0.5 \times 54 = 27 \) m/s and solve for time \( t \) using the velocity expression: \( 27 = \frac{9.8}{k}(1 - e^{-kt}) \). Using \( k = 0.1815 \text{ s}^{-1} \), solve: \( 27 = 54(1 - e^{-0.1815t}) \). This simplifies to: \( 0.5 = 1 - e^{-0.1815t} \), yielding: \( e^{-0.1815t} = 0.5 \). Taking the natural logarithm of both sides: \( -0.1815t = \text{ln}(0.5) \), thus \( t \approx \frac{\text{ln}(2)}{0.1815} \approx 3.82 \text{ seconds} \).

Key Concepts

skydiver velocity

When a skydiver jumps from a plane, her velocity changes due to two main forces: gravity and wind resistance. Gravity pulls her downward, causing an increase in speed. This force can be quantified as gravitational acceleration, approximately 9.8 meters per second squared. But wind resistance pushes against her movement, reducing her speed at a rate proportional to her current velocity. The balance between these forces determines her overall velocity at any point in time.

gravitational acceleration

Gravitational acceleration is one of the simplest concepts in physics. It refers to the constant rate at which objects accelerate when falling towards the Earth due to gravity. The value of gravitational acceleration (g) is approximately 9.8 meters per second squared (m/s²) at the Earth's surface. This means that every second, the speed of a freely falling object will increase by 9.8 m/s. For the skydiver, this acceleration acts downward, continuously increasing her velocity unless other forces, like wind resistance, act upon her.

wind resistance

Wind resistance, also known as air resistance, plays a significant role when a skydiver is falling. Unlike gravitational acceleration, which is constant, wind resistance depends on the skydiver’s velocity. It opposes the skydiver’s movement and increases with velocity. The higher the speed, the greater the wind resistance. In mathematical terms, it is often modeled as being proportional to the velocity: -kv, where k is a constant of proportionality. This means that as the skydiver's speed increases, the wind resistance increases, which eventually reduces the rate at which her velocity increases.

equilibrium solution

An equilibrium solution in this context is the velocity at which the forces of gravity and wind resistance balance out. Mathematically, this is when the rate of change of velocity is zero. By setting the right-hand side of the differential equation to zero (\( \frac{dv}{dt} = g - kv = 0 \)), we find the equilibrium solution: \( v = \frac{g}{k} \). At this velocity, the skydiver’s speed no longer changes because the downward force from gravity is perfectly balanced by the upward force from wind resistance. This equilibrium velocity is also known as the terminal velocity.

terminal velocity

Terminal velocity is the constant speed that a skydiver eventually reaches when falling. At terminal velocity, the force of gravity pulling the skydiver down is balanced by the wind resistance pushing her up. This results in zero net acceleration. For a typical skydiver, terminal velocity is around 54 meters per second. Using our equilibrium formula \( v = \frac{g}{k} \), we can determine the constant of proportionality k if we know the terminal velocity. Solving for a terminal velocity of 54 m/s, we find that \( k \approx 0.1815 \text { s}^{-1} \). This shows how k varies based on the skydiver's shape, mass, and falling position, as these affect wind resistance.