Problem 82 - Algebra and Trigonometry

Question

The velocity of a sky diver \(t\) seconds after jumping is given by \(v(t)=80\left(1-e^{-0.2 t}\right) .\) After how many seconds is the velocity 70 \(\mathrm{ft} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

The velocity is 70 ft/s after approximately 10.4 seconds.

Step 1: Understand the Problem

We are given the velocity function of a skydiver, expressed as \(v(t) = 80(1-e^{-0.2t})\). We need to find the time \(t\) when the velocity \(v(t)\) is 70 ft/s.

Step 2: Set Up the Equation

Substitute 70 ft/s for \(v(t)\) in the velocity function: \[70 = 80(1-e^{-0.2t}).\] This equation will help us find the value of \(t\).

Step 3: Isolate the Exponential Term

Subtract 80 from both sides of the equation to isolate the exponential term:\[70 = 80 - 80e^{-0.2t}.\]This simplifies to:\[80e^{-0.2t} = 80 - 70.\] \[80e^{-0.2t} = 10.\]

Step 4: Solve for the Exponential

Divide both sides of the equation by 80 to solve for the exponential:\[e^{-0.2t} = \frac{10}{80}.\] Simplifying gives: \[e^{-0.2t} = 0.125.\]

Step 5: Take the Natural Logarithm

Apply the natural logarithm to both sides of the equation to solve for \(t\):\[\ln(e^{-0.2t}) = \ln(0.125).\]This simplifies to:\[-0.2t = \ln(0.125).\]

Step 6: Solve for t

Divide by \(-0.2\) to find \(t\):\[t = \frac{\ln(0.125)}{-0.2}.\]Calculate \( \ln(0.125) \approx -2.07944 \), so:\[t \approx \frac{-2.07944}{-0.2} = 10.3972.\] Therefore, \(t\) is approximately 10.4 seconds.

Key Concepts

Understanding Velocity in Exponential Functions

Velocity is a measure of how fast an object is moving in a particular direction. When dealing with equations in physics, we often come across various mathematical representations of velocity. In the case of our skydiver, the velocity function is given as:

• \( v(t) = 80(1-e^{-0.2t}) \)

This equation shows that the velocity depends on time \( t \) and follows an exponential decay model.
Exponential functions like these are frequently used to describe growth or decay processes. Here, as the skydiver jumps, the initial acceleration decreases exponentially until reaching a terminal velocity.
Keep in mind that the expression \( 1-e^{-0.2t} \) signifies the gradual complete approach to a certain speed, in this case, 80 ft/s. Translating this function to a real-world scenario, the skydiver's speed slows its rate of increase until it stabilizes, due to air resistance.

Solving Exponential Equations

Solving equations involving exponential functions might seem challenging, but it becomes manageable by breaking down each step.
To find when the velocity \( v(t) \) is 70 ft/s, we set up the equation as follows:

• \( 70 = 80(1-e^{-0.2t}) \)

Our goal is to isolate the term with the exponential component. Subtract what isn't needed and simplify the equation:

• First, rewrite the equation \( 80e^{-0.2t} = 10 \) by rearranging and simplifying terms.
• Next, divide both sides by 80, yielding \( e^{-0.2t} = 0.125 \).

The point is to get the exponential part by itself, which simplifies thinking about how to proceed with calculations. Once the exponential term is isolated, it is ready for the next step: applying the natural logarithm.

Using the Natural Logarithm to Solve the Equation

The natural logarithm, denoted as \( \ln \), is the inverse function of the exponential function. It helps solve equations where the variable is in an exponent. This property becomes useful once you've isolated the exponential term as in \( e^{-0.2t} = 0.125 \).
By taking the natural logarithm of both sides, you can simplify the expression considerably:

• \( \ln(e^{-0.2t}) = \ln(0.125) \)
• This simplifies to \( -0.2t = \ln(0.125) \)

The great thing about logarithms is that they bring the exponent down as a coefficient, allowing us to "solve for" or isolate \( t \). Simply divide through by \(-0.2\) to find \( t \):

• \( t = \frac{\ln(0.125)}{-0.2} \)
• Calculating further, \( \ln(0.125) \approx -2.07944 \) leads to \( t \approx 10.4 \) seconds.

This method transforms the problem from an abstract function to a straightforward arithmetic calculation, solving for time relatively easily.