Problem 36 - Advanced Engineering Mathematics

Question

Assuming that the air resistance is proportional to velocity and the positive direction is downward with \(s(0)=0\) the model for the velocity is \(m d v / d t=m g-k v .\) Using separation of variables to solve this differential equation, we obtain \(v(t)=m g / k+c e^{-k t / m} .\) Then, using \(v(0)=0,\) we get \(v(t)=(m g / k)\left(1-e^{-k t / m}\right)\) Letting \(k=0.5, m=(125+35) / 32=5,\) and \(g=32,\) we have \(v(t)=320\left(1-e^{-0.1 t}\right) .\) Integrating we find \(s(t)=320 t+3200 e^{-0.1 t}+c_{1} . \quad\) Solving \(s(0)=0\) for \(c_{1}\) we find \(c_{1}=-3200,\) therefore \(s(t)=\) \(320 t+3200 e^{-0.1 t}-3200 .\) At \(t=15,\) when the parachute opens, \(v(15)=248.598\) and \(s(15)=2314.02\) At this time the value of \(k\) changes to \(k=10\) and the new initial velocity is \(v_{0}=248.598 .\) With the parachute open, the skydiver's velocity is \(v_{p}(t)=m g / k+c_{2} e^{-k t / m},\) where \(t\) is reset to 0 when the parachute opens. Letting \(m=5, g=32,\) and \(k=10,\) this gives \(v_{p}(t)=16+c_{2} e^{-2 t} .\) From \(v(0)=248.598\) we find \(c_{2}=232.598\) so \(v_{p}(t)=16+232.598 e^{-2 t} .\) Integrating, we get \(s_{p}(t)=16 t-116.299 e^{-2 t}+c_{3} .\) Solving \(s_{p}(0)=0\) for \(c_{3}\) we find \(c_{3}=116.299,\) so \(s_{p}(t)=16 t-116.299 e^{-2 t}+116.299 .\) Twenty seconds after leaving the plane is five seconds after the parachute opens. The skydiver's velocity at this time is \(v_{p}(5)=16.0106 \mathrm{ft} / \mathrm{s}\) and she has fallen a total of \(s(15)+s_{p}(5)=2314.02+196.294=2510.31 \mathrm{ft} .\) Her terminal velocity is \(\lim _{t \rightarrow \infty} v_{p}(t)=16, \mathrm{s}\) she has very nearly reached her terminal velocity five seconds after the parachute opens. When the parachute opens, the distance to the ground is \(15,000-s(15)=15,000-2,314=12,686\) ft. Solving \(s_{p}(t)=12,686\) we get \(t=785.6 \mathrm{s}=13.1 \mathrm{min} .\) Thus, it will take her approximately 13.1 minutes to reach the ground after her parachute has opened and a total of \((785.6+15) / 60=13.34\) minutes after she exits the plane.

Step-by-Step Solution

Answer

13.34 minutes total fall time after exiting the plane.

Step 1: Understand the Initial Conditions and Equations

We start with the differential equation for velocity: \( \frac{m dv}{dt} = mg - kv \). Here, the initial conditions given are \( s(0) = 0 \) and \( v(0) = 0 \). The constants provided are \( m = 5 \), \( g = 32 \), and initial \( k = 0.5 \). We'll solve this using separation of variables.

Step 2: Solve for Velocity with Initial Conditions

Using separation of variables: divide and integrate to find \( v(t) = \frac{mg}{k} (1 - e^{-\frac{kt}{m}}) \). Substitute \( k = 0.5 \), \( m = 5 \), and \( g = 32 \) to get \( v(t) = 320(1-e^{-0.1t}) \).

Step 3: Determine the Position Expression (Before Parachute Opens)

Integrate the velocity function to find the position \( s(t) = 320t + 3200e^{-0.1t} + c_1 \). Using the condition \( s(0) = 0 \), solve for \( c_1 = -3200 \), giving \( s(t) = 320t + 3200e^{-0.1t} - 3200 \).

Step 4: Adjust for Parachute Opening Conditions

At \( t = 15 \), find \( v(15) \approx 248.598 \) ft/s and \( s(15) \approx 2314.02 \) ft. Set \( k = 10 \) to simulate parachute conditions, reset time, and find \( v_p(t) = 16 + c_2 e^{-2t} \). With \( v(0) = 248.598 \), solve for \( c_2 = 232.598 \).

Step 5: Solve for Position with Parachute Open

Integrate the parachute velocity function to find \( s_p(t) = 16t - 116.299e^{-2t} + c_3 \). Using \( s_p(0) = 0 \), solve for \( c_3 = 116.299 \), giving \( s_p(t) = 16t - 116.299e^{-2t} + 116.299 \).

Step 6: Calculate Total Fall Distance at Final Time Check

Twenty seconds after jump, or five seconds after parachute opens, evaluate \( v_p(5) \approx 16.0106 \) ft/s. Total distance fallen is \( s(15) + s_p(5) = 2314.02 + 196.294 = 2510.31 \) ft.

Step 7: Determine Time Until Ground Impacts

To find when the skydiver reaches the ground: calculate how long after parachute opens for \( s_p(t) = 12,686 \) leading to \( t \approx 785.6 \) seconds. Adding initial 15 seconds gives total fall time \( 785.6 + 15 = 800.6 \) seconds, or \( 13.34 \) minutes.

Key Concepts

Separation of Variables

Separation of variables is a useful technique in solving differential equations, especially when they can be written in a form where the variables are isolated on different sides of the equation. Here, we started with the equation \( \frac{m dv}{dt} = mg - kv \). Our goal was to find an expression for the velocity \( v(t) \).
To apply separation of variables, rearrange the terms to isolate the variables. We can write \( \frac{dv}{mg - kv} = \frac{dt}{m} \). Now each side depends on a single variable: all terms with \( v \) are on one side, and all terms with \( t \) are on the other.
This allows us to integrate both sides separately:

• The left-hand side integrates with respect to \( v \).
• The right-hand side integrates with respect to \( t \).

After integration, we solve for \( v(t) \), resulting in \( v(t) = \frac{mg}{k}(1 - e^{-\frac{kt}{m}}) \). This method simplifies differential equations by breaking them down step by step, making them easier to solve.

Initial Conditions

Initial conditions are essential for finding the exact solution to a differential equation, as they allow for determining any constants of integration. In this problem, the initial conditions given were \( s(0) = 0 \) and \( v(0) = 0 \). These conditions reflect the starting point of the problem.
When we solved the velocity equation, the initial condition \( v(0) = 0 \) was crucial for finding the constant \( c \). Using \( v(0) = 0 \), we solved for \( c \) in the expression \( v(t) = \frac{mg}{k} + ce^{-\frac{kt}{m}} \), ensuring the velocity correctly starts from zero.
Similarly, we used \( s(0) = 0 \) to solve for the constant \( c_1 \) after integrating the velocity function to obtain a position function \( s(t) = 320t + 3200e^{-0.1t} - 3200 \).
Without initial conditions, we wouldn't have a unique solution corresponding to the physical setup of the problem, like the skydiver's initial rest at the start of the jump.

Terminal Velocity

Terminal velocity is a key concept when analyzing objects falling through a resistive force like air. It occurs when the force of gravity pulling the object downward is balanced by the air resistance opposing its motion. At this point, the net force is zero, and the velocity becomes constant.
In our scenario, the terminal velocity can be found by letting \( t \to \infty \) in the velocity expression. For example, before the parachute opens with \( k = 0.5 \), the terminal velocity is \( \frac{mg}{k} = 320 \text{ ft/s} \). This is calculated by setting the exponential term to zero, as \( e^{-\frac{kt}{m}} \) approaches zero.
When the parachute opens, increasing the drag constant \( k \) to 10, the terminal velocity significantly decreases. The new terminal velocity becomes \( \frac{mg}{k} = 16 \text{ ft/s} \), indicating the parachute's effectiveness in slowing down the descent.
Understanding terminal velocity is vital in predicting how quickly an object will eventually move when dropped through a fluid, and how devices like parachutes work to alter the rate of descent.