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When a skydiver jumps from an airplane, her downward velocity, in feet per second, before she opens her parachute is given by \(v=176\left(1-0.834^{t}\right)\), where \(t\) is the number of seconds that have elapsed since she jumped from the airplane. We found earlier that the terminal velocity for the skydiver is 176 feet per second. How long does it take to reach \(90 \%\) of terminal velocity?

Short Answer

Expert verified
It takes about 12.63 seconds to reach 90% of terminal velocity.

Step by step solution

01

Identify Terminal Velocity and the Desired Velocity

The terminal velocity given in the problem is 176 feet per second. We want to find when the velocity is 90% of the terminal velocity. Calculate 90% of 176 feet per second: \( v = 0.9 \times 176 = 158.4 \) feet per second.
02

Set Up the Equation

We know the velocity function is given by \( v = 176(1 - 0.834^t) \). We need to find the time \( t \) when the velocity \( v = 158.4 \). Set up the equation \( 176(1 - 0.834^t) = 158.4 \).
03

Solve for 1 - 0.834^t

Divide both sides of the equation by 176 to isolate \( 1 - 0.834^t \):\( 1 - 0.834^t = \frac{158.4}{176} \).Calculate the fraction \( 1 - 0.834^t = 0.9 \).
04

Solve for 0.834^t

Subtract 0.9 from both sides to isolate the term involving \( t \):\( 0.834^t = 1 - 0.9 = 0.1 \).
05

Use Logarithms to Solve for t

Now, apply the natural logarithm to both sides to solve for \( t \):\( \ln(0.834^t) = \ln(0.1) \).Using the logarithmic identity \( \ln(a^b) = b\ln(a) \), this becomes:\( t \ln(0.834) = \ln(0.1) \).Divide both sides by \( \ln(0.834) \):\( t = \frac{\ln(0.1)}{\ln(0.834)} \).
06

Calculate the Time t

Calculate the value of \( t \) using a calculator:\( t \approx \frac{-2.3026}{-0.1823} \approx 12.63 \). Thus, it takes approximately 12.63 seconds to reach 90% of terminal velocity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are a type of mathematical expression where a base number is raised to a power or an exponent. In the context of velocity problems, exponential functions are crucial, as they help model situations where a quantity changes at a rate proportional to its current value. This is often seen in natural processes like population growth or radioactive decay.

For the given exercise, the velocity of a skydiver is defined by the exponential function: \[ v = 176 (1 - 0.834^t) \].
This expression shows how the velocity increases over time until it approaches a maximum or terminal velocity as determined by the equation.

Key characteristics of exponential functions include:
  • Rapid change, either growing or decreasing, depending on the base of the exponent
  • A base less than one signifies exponential decay, while a base greater than one implies exponential growth
  • They approach a limit over time, like terminal velocity in this scenario.
Understanding exponential functions is fundamental when dealing with velocity equations in physics, as they describe how quickly an object moves or ceases to accelerate.
Terminal Velocity
Terminal velocity is a critical concept in the study of motion and physics. It refers to the constant speed that a freely falling object eventually reaches when the resistance of the medium (like air) prevents further acceleration. When an object achieves terminal velocity, the net force acting on it is zero, meaning that the force of gravity is balanced by the drag force of the medium.

In the given problem, the skydiver's terminal velocity is specified as 176 feet per second. This is the highest velocity the skydiver will reach before opening the parachute, and it signifies that the air resistance has equaled the gravitational pull on the skydiver.

Key factors influencing terminal velocity include:
  • The mass of the falling object
  • The surface area of the object
  • The density of the medium it is moving through
In this exercise, understanding terminal velocity helps in determining how long it takes for the skydiver to reach a significant portion of that speed, as well as the limitations on her movement due to drag forces.
Logarithms
Logarithms are mathematical tools used to solve equations involving exponential expressions. They help to "undo" exponential functions, making it easier to isolate variables that are part of exponentials. In simple terms, a logarithm tells us how many times a particular number, called the base, is multiplied to achieve another number.

In the context of this exercise, we use logarithms to solve for the time \( t \) in the equation \( 0.834^t = 0.1 \). By applying the logarithm, this exponential equation is transformed into a more manageable algebraic equation:

\( \ln(0.834^t) = \ln(0.1) \).

This transformation facilitates solving for the time variable \( t \) by using the identity \( \ln(a^b) = b\ln(a) \).

Some important aspects of logarithms include:
  • Inverse operations of exponents
  • Allowing operations involving very small or large numbers to be simplified and solved
  • Commonly used bases include 10 (common logarithm) or e (natural logarithm), where e is approximately 2.71828.
Mastery of logarithms is essential in tackling exponential functions encountered in math and physics.
Algebraic Equations
An algebraic equation involves expressions containing numbers, variables, and operations like addition, subtraction, multiplication, and division. Solving these equations is about finding the value of the unknowns that make the equation true. Algebraic equations form the backbone of solving real-world problems, including those found in physics and engineering.

In the skydiver problem, arranging and solving equations step-by-step was key to finding the time it takes to reach 90% of terminal velocity. Initially, the problem provided an equation: \[ 176 (1 - 0.834^t) = 158.4 \].

Algebraic manipulation was used to simplify and rearrange terms:
  • Isolating terms to solve for exponential parts
  • Dividing by constants to simplify the equation
  • Applying logarithms to solve for the exponent
Understanding how to manipulate and solve such equations is fundamental, as it allows students to interpret and solve practical problems involving physics, like determining different phases of motion in the case of a skydiver.

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Most popular questions from this chapter

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