Problem 28 - Precalculus Student Solutions Manual 5th

Question

Wind Speed Joe traveled against the wind in a small plane for 3 hr. The return trip with the wind took 2.8 hr. Find the speed of the wind if the speed of the plane in still air is \(180 \mathrm{mph}\)

Step-by-Step Solution

Answer

The wind speed is approximately \(6.21 \text{ mph}\).

Step 1: Define Variables

Let the speed of the wind be represented by the variable \(w\). The speed of the plane in still air is given as \(180 \text{ mph}\).

Step 2: Setup Equations for Each Trip

For the trip against the wind: \( \text{Speed} = 180 - w \). For the return trip with the wind: \( \text{Speed} = 180 + w \).

Step 3: Express Distances Traveled

The distance Joe travels against the wind is \(3 \text{ hours} \times (180 - w) \text{ mph} = 3(180 - w)\). The distance Joe travels with the wind is \(2.8 \text{ hours} \times (180 + w) \text{ mph} = 2.8(180 + w)\). Since both distances are the same, we equate them: \(3(180 - w) = 2.8(180 + w)\).

Step 4: Solve the Equation

Expand and solve for \(w\): \(3(180 - w) = 2.8(180 + w)\) which expands to \(540 - 3w = 504 + 2.8w\). Combine like terms: \(540 - 504 = 2.8w + 3w\) leading to \(36 = 5.8w\). Finally, solve for \(w\): \(w = \frac{36}{5.8} \approx 6.21 \text{ mph}\).

Key Concepts

distance equals rate times time

In precalculus, a key formula you'll frequently encounter is \( \text{distance} = \text{rate} \times \text{time} \). This means the distance traveled equals the speed (or rate) of travel multiplied by the time spent traveling.

For instance, if you drive a car at a speed of 60 mph for 2 hours, the distance covered is calculated by:
\[ \text{Distance} = 60 \text{ mph} \times 2 \text{ hr} = 120 \text{ miles} \].

This simple yet powerful formula underpins many problems in physics and mathematics, including the one we're exploring about Joe's airplane trip against and with the wind.

By setting up this formula correctly, we can solve for various unknowns, such as speed, time, or distance, as long as the other two variables are known. In our exercise, Joe's distance traveled against and with the wind was assumed equal, allowing us to set up an equation and solve for the wind's speed.

solving linear equations

Solving linear equations is a fundamental aspect of solving word problems in precalculus. A linear equation generally takes the form \( ax + b = c \), where \( a \), \( b \), and \( c \) are constants.

Let's consider Joe's problem with wind speed. We started with the equation:
\[3(180 - w) = 2.8(180 + w)\].

To solve this, we first expanded both sides:
\[540 - 3w = 504 + 2.8w\].

Next, we rearranged the terms to combine like terms:
\[540 - 504 = 2.8w + 3w\], leading us to
\[36 = 5.8w\].

Finally, we solved for \( w \) by dividing both sides by 5.8:
\[w = \frac{36}{5.8} \approx 6.21 \text{ mph}\].

By following these steps carefully, you can solve similar linear equations in any word problem scenario.

variables in word problems

Variables are symbols used to represent unknown quantities in math problems. They allow us to set up equations and solve for these unknown values.

In Joe's wind speed problem, we used the variable \( w \) to represent the speed of the wind.

When approaching word problems, it's crucial to:

• Identify all the given quantities.
• Determine what you need to find.
• Assign appropriate variables to the unknowns.

For example, we know the plane's speed in still air (180 mph) and the time for each trip (3 hours against the wind and 2.8 hours with the wind).

Defining the wind speed as \( w \) allowed us to set up the relationship between the plane's speed against and with the wind:
\[\text{Speed against wind} = 180 - w\]\[\text{Speed with wind} = 180 + w\].

After defining the variables, we created the equations based on the distances, solved the resulting linear equation, and found the wind speed.
Understanding how to define and use variables is a critical skill in tackling word problems effectively.