Problem 47 - Algebra and Trigonometry

Question

A man flies a small airplane from Fargo to Bismarck, North Dakota—a distance of 180 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

The airplane's speed in still air is 105 mph, and the wind speed is 15 mph.

Step 1: Define Variables

Let the speed of the airplane in still air be \( a \) mph, and the speed of the wind be \( w \) mph. The speed while flying to Bismarck (against the wind) is \( a - w \) mph, and the speed flying back to Fargo (with the wind) is \( a + w \) mph.

Step 2: Set Up Equations for Each Trip

For the trip to Bismarck:\[\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{180}{a-w} = 2 \text{ (hours)}\]This simplifies to:\[a - w = 90 \text{ (Equation 1)}\]For the return trip to Fargo:\[\text{Time} = \frac{180}{a+w} = \frac{3}{2} \text{ (hours)}\]This simplifies to:\[a + w = 120 \text{ (Equation 2)}\]

Step 3: Solve the System of Equations

Add Equation 1 and Equation 2:\[(a - w) + (a + w) = 90 + 120\]This simplifies to:\[2a = 210\]Divide by 2:\[a = 105\]Substitute \(a = 105\) back into Equation 1:\[105 - w = 90\]Solve for \(w\):\[w = 15\]

Step 4: Interpret the Results

The speed of the airplane in still air is 105 mph, and the speed of the wind is 15 mph.

Key Concepts

Solving Systems of Equations

Solving systems of equations is a fundamental aspect of algebra that allows us to find the values of variables that satisfy multiple mathematical statements simultaneously. In this exercise, we encounter two equations, each representing one leg of the airplane trip.
The equations are derived from the relationship between distance, speed, and time. To tackle the problem, we need to solve these equations together as a system.

• Equation 1: Derived from the trip to Bismarck, it states that the effective speed is the airplane's speed minus the wind speed.
• Equation 2: Originates from the return trip and represents the airplane's speed plus the wind's speed.

Two main methods to solve a system like this are substitution and addition (elimination). In our case, addition is used to eliminate one of the variables (wind speed), making it easier to find the airplane's speed. Once we find the speed of the airplane in still air, substituting back into one of the original equations will provide the wind speed. This step-by-step solving ensures accuracy and clarity, providing a solid foundation for understanding systems of equations.

Distance-Speed-Time Relationship

The relationship between distance, speed, and time is crucial for solving problems involving motion. It is often expressed with the formula: \[\text{Distance} = \text{Speed} \times \text{Time}\]In the given exercise, this relationship helps in forming the key equations.

• For the outbound trip to Bismarck, the headwind slows the plane down, so the effective speed becomes \(a - w\), expressed through the time formula as \(\frac{180}{a-w} = 2\).
• On the return trip to Fargo, the plane benefits from the tailwind, speeding up the effective rate, expressed as \(\frac{180}{a+w} = \frac{3}{2}\).

Understanding this relationship is essential because it offers a way to convert real-world problems into solvable algebraic equations. Knowing how to manipulate this formula allows us to adjust for factors such as opposing or aiding forces (wind, in this case), thus facilitating the solution of travel-time problems.

Algebraic Problem-Solving

Algebraic problem-solving is about converting verbal or contextual problems into mathematical equations and then manipulating these equations to find solutions. This exercise is a prime example of translating a real-world scenario into algebraic form.
The key steps are:

• Define Variables: This involves assigning letters like \(a\) for airplane speed and \(w\) for wind speed to unknown quantities.
• Form Equations: Using the distance-speed-time relationship, we create equations that represent the situation.
• Manipulate and Solve: By using addition or substitution methods, solve these equations systematically.

The methodical approach of understanding what each variable represents and how they're linked through algebraic expressions is vital. It's not just about finding numbers but comprehending the interplay of different elements (like wind speed) that affect the problem scenario. This makes algebra a powerful tool for problem-solving in real-life contexts.