Question

It took 3 hours for a plane, flying against the wind, to travel 900 miles from Alabama to Minnesota. The "ground speed" of the plane is 300 miles per hour. On the return trip, the flight took only 2 hours with a ground speed of 450 miles per hour. During both flights the speed and the direction of the wind were the same. The plane's speed decreases or increases because of the wind as the verbal model below shows. Speed in still air \(]-[\text { Wind speed }]=\) Ground speed against wind Speed in still air \(+\text { Wind speed }]=\) Ground speed with wind Assign labels to the verbal model shown above. Use the labels to translate the verbal model into a system of linear equations.

Short answer

The speed of the plane in still air is 375 miles per hour and the wind speed is 75 miles per hour.

Step-by-step solution

Assign Variables to Unknowns

As per the verbal model, assign variables. The 'speed in still air' can be represented by a variable 's' (measured in miles per hour) which represents the speed of the plane without any wind. The 'wind speed' can be represented by 'w' (also in miles per hour).

Translate the Information into Equations

Translate the model into a system of two equations according to the given conditions. When flying against the wind, the ground speed (300 miles per hour) equals the plane speed minus the wind speed (s - w). So, the first equation is: \(s - w = 300\). When flying with the wind, the ground speed (450 miles per hour) is the plane speed plus the wind speed (s + w). So, the second equation is: \(s + w = 450\).

Solve the system of equations

Add the two equations: \(s - w + s + w = 300 + 450\). This simplifies to \(2s = 750\). Solve this equation for \(s\) to find the speed of the plane in still air: \(s = 375\) miles per hour. Substitute \(s = 375\) into the first equation, \(s - w = 300\), to find the wind speed: \(w = 375 - 300 = 75\) miles per hour.

Key concepts

Algebraic Modeling

Algebraic modeling is a powerful tool used in mathematics to represent real-world situations through mathematical expressions and equations. By using variables and algebraic operations, we create models that can replicate and predict various scenarios.

For example, consider the challenge of finding the speed of an airplane in still air and the speed of the wind, given the ground speeds with and against the wind. Here, algebraic modeling translates the word problem into a verbal model first and then assigns variables to unknown quantities, such as 's' for the airplane's speed in still air and 'w' for wind speed.

Once the variables are defined, we use the information provided to build a system of linear equations: In our case,

• When flying against the wind: Ground speed = Speed in still air - Wind speed
• When flying with the wind: Ground speed = Speed in still air + Wind speed

The ground speed against the wind is given as 300 miles per hour, and with the wind is 450 miles per hour. These equations encapsulate the relationship between the quantities involved and are ready to be solved, showcasing the power of algebraic modeling in turning real-life problems into solvable mathematical tasks.

Variables in Algebra

In algebra, variables are symbols that represent unknown values. They are essential in formulating algebraic expressions, equations, and functions. Variables allow us to generalize problems, making it possible to solve not just one specific case, but a whole class of problems that follow the same pattern.

Returning to our airplane scenario, we've defined two variables: 's' for the speed in still air and 'w' for the wind speed. These variables are placeholders for numbers we don't yet know but are determined to find. Understanding how to effectively choose and use variables is a foundational skill in algebra. It's often helpful to choose variable names that relate to the concept they represent, making the problem easier to interpret.

For instance, using 's' to represent the speed of the airplane makes it intuitive when we say that the ground speed against the wind is represented by the expression 's - w', and when assisted by the wind, by 's + w'. Through this method, the abstract nature of algebra becomes tied to tangible real-world quantities.

Solving Linear Systems

Solving linear systems is the process of finding the values of variables that satisfy multiple linear equations simultaneously. When equations are crafted to represent real-world situations, solving these systems yields solutions to problems, just like the speed of an airplane and the wind speed in our example.

To solve a system of linear equations, we generally use methods such as substitution, elimination, or graphing. In our airplane case, we have the following system of equations:

Equations for the Airplane Problem

• s - w = 300 (against the wind)
• s + w = 450 (with the wind)

The elimination method works perfectly here by adding the two equations, effectively eliminating 'w' and allowing us to solve for 's'. Once we know 's', we can substitute it back into either equation to find 'w'. The solutions to these equations give us valuable information; they indicate the aircraft's true speed without the influence of wind and the wind's speed itself.

The method chosen depends on the particular system at hand, but whichever path is taken, the goal remains the same: to determine the exact values that make all the equations true at the same time. This highlights the fascinating aspect of linear systems; they can model and decipher complexities in our world, making algebra a powerful tool in various fields beyond pure mathematics.