Problem 91 - Intermediate Algebra

Question

Sky Diving. Before the parachute opens, a skydiver's velocity in meters per second is modeled by the function \(f(t)=50\left(1-e^{-0.2 t}\right)\) where \(f(t)\) is the velocity at time \(t .\) Find the velocity after 20 seconds of free fall.

Step-by-Step Solution

Answer

The velocity after 20 seconds is approximately 49.085 m/s.

Step 1: Understand the Problem

We are given a function \( f(t) = 50(1 - e^{-0.2t}) \) that models the skydiver's velocity in meters per second. We need to find the velocity after 20 seconds, which means we need to evaluate this function at \( t = 20 \).

Step 2: Substitute the Value of Time

Substitute \( t = 20 \) into the function. This gives us \( f(20) = 50 \left(1 - e^{-0.2 \times 20}\right) \).

Step 3: Simplify the Exponent

Calculate the exponent: \(-0.2 \times 20 = -4\). This simplifies the expression to \( f(20) = 50 \left(1 - e^{-4}\right) \).

Step 4: Evaluate the Exponential Term

Calculate \( e^{-4} \). Using a calculator, find that \( e^{-4} \approx 0.0183 \).

Step 5: Simplify the Expression

Substitute \( e^{-4} \approx 0.0183 \) into the expression: \( f(20) = 50 \times (1 - 0.0183) \).

Step 6: Calculate the Final Velocity

Simplify: \( f(20) = 50 \times 0.9817 \). Compute \( 50 \times 0.9817 \approx 49.085 \).

Step 7: State the Final Result

After substituting and calculating, the velocity after 20 seconds of free fall is approximately \( 49.085 \) meters per second.

Key Concepts

Modeling Real-World Scenarios

In mathematics, one of the most powerful applications is modeling real-world scenarios using functions. These functions are valuable tools, as they help us represent complex phenomena with simple equations. The skydiver's velocity problem is a prime example.

• We use the given function \( f(t) = 50(1 - e^{-0.2t}) \) to model how the velocity evolves over time.
• This equation models the process of a skydiver accelerating until reaching terminal velocity, which is the constant speed during free fall.

By understanding the equation, we can predict the behavior of the skydiver's velocity at any moment, even if we don't observe it directly. We substitute different values for \( t \) to determine velocity at specific points in time, providing us insights into the dynamics at play.

Exponential Decay

Exponential decay describes processes where quantities decrease rapidly at first and then slow down over time. In our skydiving function, the term \( e^{-0.2t} \) represents this concept.

• As time \( t \) increases, the exponent \(-0.2t\) becomes more negative, making \( e^{-0.2t} \) smaller, approaching zero.
• This implies that the difference \( 1 - e^{-0.2t} \) grows closer to 1, indicating that velocity is increasing.

The constant 0.2 in \(-0.2t\) controls the rate of decay. A larger constant would mean a faster approach to terminal velocity, while a smaller one would take more time. Recognizing how these changes in the exponential term affect the function helps in understanding how exponential decay works in various real-world scenarios.

Function Evaluation

Function evaluation is the process of determining the value of a function given a specific input. It’s one of the fundamental tasks when working with functions.

• In the given scenario, you're asked to find the skydiver's velocity after 20 seconds, which means substituting \( t = 20 \) into the function \( f(t) \).
• Replace the variable \( t \) in the function \( f(t) = 50(1 - e^{-0.2t}) \) with 20: \( f(20) = 50(1 - e^{-4}) \).

This part of the process involves straightforward substitution, simplifying expressions, and possibly using a calculator to handle exponential terms like \( e^{-4} \). Function evaluation allows you to generate specific outputs from a function, thereby predicting or analyzing real-world phenomena accurately.