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The problem presented is an initial value problem involving a differential equation modeling the velocity of a person who has bailed out of a helicopter with a parachute, taking into account the effect of air resistance. 

The equation given is:

\[
\frac{dv}{dt} = 32 - 1.76v, \quad v(0) = 0
\]

where:
- \( v(t) \) is the velocity in feet per second (ft/s),
- \( t \) is the time in seconds,
- The constant \( k = 1.76 \) represents the air resistance proportionality constant,
- The term 32 represents the acceleration due to gravity in ft/s².

### Part 1: Limiting Velocity
The limiting velocity is the velocity that the person will approach as time \( t \) becomes large. This occurs when the velocity stops changing, i.e., when \( \frac{dv}{dt} = 0 \).

Solving for \( v \) when \( \frac{dv}{dt} = 0 \):

\[
0 = 32 - 1.76v
\]
\[
v = \frac{32}{1.76} = 18.18 \, \text{ft/s}
\]

So, the limiting velocity is 18.18 ft/s.

### Part 2: Time to Reach 95% of Limiting Velocity
Next, we need to determine how long it will take for the person to reach 95% of the limiting velocity. Let’s solve the differential equation to get the velocity as a function of time.

### Solving the Differential Equation
The equation is separable, so we can solve it by separating variables:

\[
\frac{dv}{32 - 1.76v} = dt
\]

Integrating both sides:

\[
\int \frac{1}{32 - 1.76v} \, dv = \int 1 \, dt
\]

The left-hand side is a standard logarithmic integral:

\[
-\frac{1}{1.76} \ln |32 - 1.76v| = t + C
\]

Exponentiating both sides and solving for \( v(t) \), we obtain the velocity as a function of time. To get the particular solution, we apply the initial condition \( v(0) = 0 \). From this, we can find the time it takes to reach 95% of the limiting velocity, which is \( 0.95 \times 18.18 = 17.27 \) ft/s.

Let me calculate the exact time it takes to reach 95% of the limiting velocity.The real solution for the time it takes to reach 95% of the limiting velocity is approximately:

\[
t = 1.70 \, \text{seconds}
\]

This means it will take about 1.7 seconds to reach 95% of the limiting velocity (which is approximately 17.27 ft/s).

### Recap and Further Questions:
1. **Limiting Velocity:** The limiting velocity is 18.18 ft/s.
2. **Time to 95% Limiting Velocity:** You will reach 95% of the limiting velocity (17.27 ft/s) in approximately 1.7 seconds.

Would you like more details on any step? Here are some further questions you might explore:

1. What happens if the air resistance constant \( k \) increases?
2. How does the initial condition \( v(0) = 0 \) affect the solution?
3. Could you find the time to reach other percentages of the limiting velocity, like 99%?
4. What would be the behavior of the system if there was no air resistance?
5. How would a different gravitational constant, such as on another planet, impact the limiting velocity?

### Tip:
When dealing with separable differential equations, always remember to isolate the variables and integrate both sides step by step carefully.

You bail out of a helicopter and pull the ripcord of your parachute. Now the air resistance proportionality constant is k = 1.76, so your downward velocity satisfies the initial value problem below, where v is measured in ft/s and t in seconds. In order to investigate your chances of survival, construct a slope field for this differential equation and sketch the appropriate solution curve. What will your limiting velocity be? Will a strategically located haystack do any good? How long will it take you to reach 95% of your limiting velocity? dv/dt = 32 - 1.76v, v(0) = 0

Learn how to solve a parachute drop velocity problem using differential equations. This problem involves calculating the limiting velocity and determining the time to reach 95% of the limiting velocity with a constant air resistance factor. Suitable for students in Grades 11-12 and undergraduate courses.

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