Problem 130 - College Algebra with Modeling and Visualization

Question

A plane flies 1500 miles against the wind in 3 hours and 45 minutes. The return trip with the wind takes 3 hours. Assume that the wind speed stays constant. Find the speed of the wind and the speed of the airplane with no wind.

Step-by-Step Solution

Answer

The airplane's speed is 450 mph, and the wind speed is 50 mph.

Step 1: Convert Time to Hours

The given time for the first trip is 3 hours and 45 minutes. First, convert the 45 minutes into hours. Since there are 60 minutes in an hour, 45 minutes is equivalent to \( \frac{45}{60} = 0.75 \) hours. Therefore, the total time for the first trip is \( 3.75 \) hours.

Step 2: Establish Equations for Speeds

Let \( x \) be the speed of the plane in still air and \( w \) be the speed of the wind. The plane's effective speed against the wind is \( x - w \) and with the wind is \( x + w \). According to the problem, the time against the wind is 3.75 hours for 1500 miles and with the wind is 3 hours for the same distance.

Step 3: Create Distance Equations

Using the equation \( \text{Distance} = \text{Speed} \times \text{Time} \), for the trip against the wind: \( 1500 = (x - w) \times 3.75 \). For the return trip with the wind: \( 1500 = (x + w) \times 3 \).

Step 4: Solve First Equation for \( x - w \)

From \( 1500 = (x - w) \times 3.75 \), divide both sides by 3.75 to find \( x - w \). This gives \( x - w = \frac{1500}{3.75} = 400 \).

Step 5: Solve Second Equation for \( x + w \)

From \( 1500 = (x + w) \times 3 \), divide both sides by 3 to find \( x + w \). This gives \( x + w = \frac{1500}{3} = 500 \).

Step 6: Solve System of Equations for \( x \) and \( w \)

Now solve the system of equations: \( x - w = 400 \) and \( x + w = 500 \). Add these two equations to eliminate \( w \):\( (x - w) + (x + w) = 400 + 500 \)\( 2x = 900 \)\( x = \frac{900}{2} = 450 \).Plug \( x = 450 \) back into \( x + w = 500 \):\( 450 + w = 500 \)\( w = 500 - 450 = 50 \).

Step 7: Finalize the Solution

The speed of the airplane in still air is 450 mph, and the wind speed is 50 mph.

Key Concepts

Distance-Speed-Time Problems

Distance-speed-time problems are common in algebra and help us understand relationships between different physical quantities. In these problems, we typically know some of the information, like distance and time, and need to find the missing one, usually speed. To solve them, we rely on the simple formula:

• Distance = Speed × Time

This formula is foundational because it links how far an object travels, how fast it is moving, and how long it takes to reach its destination.
For example, if a plane travels 1500 miles against the wind in 3.75 hours, you can calculate its effective speed by dividing the distance by the time.
This type of problem involves both calculating with and without additional factors, such as wind, which impacts the speed and requires adjusting the calculations accordingly.

System of Equations

In algebra, a system of equations is a set of two or more equations with the same set of variables. Solving these systems means finding values for the variables that make all the equations true simultaneously.
In our exercise, we use two equations based on distance-speed-time relationships:

• Against the wind: \(1500 = (x - w) imes 3.75\)
• With the wind: \(1500 = (x + w) imes 3\)

Here, \(x\) is the speed of the airplane in still air, and \(w\) is the wind speed.
We solve these equations simultaneously to find both variables. Systems of equations are powerful tools because they allow us to deal with complex problems involving multiple unknowns. In this case, adding or subtracting equations allows for the elimination of variables to simplify the solution.

Problem Solving Steps

Solving algebraic problems step by step ensures accuracy and clarity. Let's break down how this works using our example. First, it is crucial to ensure all given information is in suitable units, like converting minutes to hours. This sets the foundation for forming correct equations.
Next, create the equations based on relationships you identify in the problem, such as using distance-speed-time for each leg of the journey.
Once the equations are formed, solve for one variable at a time. For instance, calculate the speeds against and with the wind separately.

• Solve for \(x - w\) and \(x + w\) independently.
• Then solve the system of equations by adding them to find \(x\), and substitute back to find \(w\).
• Double-check calculations and make sure the solution makes sense within the context of the problem.

By following clear, logical steps, you can tackle complex problems efficiently and reduce the likelihood of errors.