Problem 47 - Precalculus: Mathematical for Calculus

Question

A man flies a small airplane from Fargo to Bismarck, North Dakota-a distance of \(180 \mathrm{mi}\). Because he is flying into a head wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 h 12 min. What is his speed in still air, and how fast is the wind blowing?

Step-by-Step Solution

Answer

Speed in still air: 120 mph; Wind speed: 30 mph.

Step 1: Define the Variables

Let's assume the speed of the airplane in still air is \(x\) miles per hour (mph) and the speed of the wind is \(y\) mph.

Step 2: Set Up Equations for Each Trip

For the trip from Fargo to Bismarck (against the wind), the airplane's effective speed is \(x - y\) mph and it takes 2 hours to travel 180 miles. This gives us the equation: \[ 180 = 2(x - y) \].For the return trip (with the wind), the airplane's effective speed is \(x + y\) mph and it takes 1 hour and 12 minutes, which is 1.2 hours, to travel 180 miles. This gives us the equation: \[ 180 = 1.2(x + y) \].

Step 3: Solve the Equations

First, simplify both equations. From the first equation, we have:\[ 180 = 2(x - y) \Rightarrow x - y = 90. \]From the second equation, we have:\[ 180 = 1.2(x + y) \Rightarrow x + y = 150. \]

Step 4: Add and Subtract the Simplified Equations

Add the two equations to eliminate \(y\):\[ (x - y) + (x + y) = 90 + 150 \Rightarrow 2x = 240 \Rightarrow x = 120. \]Subtract the first equation from the second equation to eliminate \(x\):\[ (x + y) - (x - y) = 150 - 90 \Rightarrow 2y = 60 \Rightarrow y = 30. \]

Step 5: Interpret the Results

The airplane's speed in still air is \(120\) mph and the wind speed is \(30\) mph.

Key Concepts

Speed and Distance Problems

Speed and distance problems often involve calculating the time, speed, or distance in various contexts. A classic example is determining travel details under varying conditions.
It revolves around understanding that the relationship between speed, distance, and time can be expressed by the formula:

• Distance = Speed × Time

This principle helps in solving problems related to moving objects like vehicles or aircrafts. For instance, if you know the distance and time, you can solve for speed by rearranging the formula:

• Speed = Distance / Time

When elements like wind or current come into play, they affect the effective speed. A headwind, which opposes motion, reduces effective speed, while a tailwind, which aids motion, increases it. In our exercise, the airplane faces a headwind while traveling to its destination and a tailwind on its return trip.
This affects the travel time for the same distance. Understanding these changes helps in setting up correct equations to find unknowns such as the aircraft's still air speed or wind speed.

Linear Equations

Linear equations are foundational in algebra and they depict a straight line when graphed on a coordinate plane.
Each equation represents a direct relationship between two variables, typically in the form:

• y = mx + b

where 'm' is the slope and 'b' is the y-intercept. In speed and distance problems involving different conditions, we can set up linear equations to find unknowns. In our example, we created equations for each leg of the trip:

• 180 = 2(x - y) for the outbound flight
• 180 = 1.2(x + y) for the return trip

These equations help solve for the variables, which in this case are the plane's speed in still air and the wind speed.
By solving these, we can illustrate the intersection and impact of different elements like headwinds and tailwinds on effective travel speed. Getting skilled at forming and solving linear equations is essential for tackling more complex algebraic problems.

Problem Solving in Algebra

Problem solving in algebra is an essential skill that involves translating real-world scenarios into mathematical models. This means identifying what is given, defining variables for unknowns, and forming equations to represent the relationships.
Once equations are set up, the next step is to simplify and solve them to find the unknowns. Here's the process generally followed:

• Define the variables: Understand what each variable represents in the context.
• Set up the equations: Use the relationships and formulae related to the problem.
• Simplify the equations: Make them easier to solve through simplification or manipulation.
• Solve the equations: Use algebraic methods like substitution or elimination.
• Interpret the solution: Understand what the solution means in context.

In our exercise, solving for the airplane's speed and the wind speed required all these steps.
By defining variables 'x' for speed in still air and 'y' for wind speed and forming equations based on given travel times and distances, we executed a systematic approach to reach the solution.