Problem 63 - Algebra and Trigonometry

Question

Airplane Speed \(A\) 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 \(\mathrm{h} 12\) 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 120 mph, and the wind speed is 30 mph.

Step 1: Define Variables

Let's define the variables needed to solve the problem. Let \( v_p \) be the speed of the airplane in still air (without wind), and let \( v_w \) be the speed of the wind. Since both trips cover the same distance of 180 miles, we will use these variables to set up equations.

Step 2: Convert Units

The return trip time is given as 1 hour and 12 minutes. First, we need to convert this time into hours. \(12\) minutes is equivalent to \(\frac{12}{60} = 0.2\) hours, so the total time for the return trip is \(1 + 0.2 = 1.2\) hours.

Step 3: Set Up Equations for Each Trip

For the trip to Bismarck, the plane heads into the wind. Therefore, the effective speed is \(v_p - v_w\). The time taken is 2 hours, so we have:\[ 180 = (v_p - v_w) imes 2 \] For the return trip, the wind is helping, so the effective speed is \(v_p + v_w\). The time taken is 1.2 hours, so:\[ 180 = (v_p + v_w) imes 1.2 \]

Step 4: Simplify and Solve the Equations

Simplify the two equations:1. \(180 = 2(v_p - v_w)\) gives \(90 = v_p - v_w\) 2. \(180 = 1.2(v_p + v_w)\) gives \(150 = v_p + v_w\)Now, solve this system of equations. Add the two equations:\[ 90 + 150 = (v_p - v_w) + (v_p + v_w) \]\[ 240 = 2v_p \]Now solve for \(v_p\):\[ v_p = \frac{240}{2} = 120 \text{ mph} \] Now solve for \(v_w\) using the equation \(90 = v_p - v_w\):\[ 90 = 120 - v_w \]\[ v_w = 120 - 90 = 30 \text{ mph} \]

Step 5: Conclude the Solution

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

Key Concepts

Systems of Equations

Understanding word problems often starts with setting up equations correctly. A **system of equations** involves two or more equations that are solved together because they share variables. In this problem, we have two unknowns: the speed of the airplane in still air, \( v_p \), and the speed of the wind, \( v_w \). We need two equations to determine these unknowns.

- First, we use the fact that the airplane covers a distance of 180 miles both ways. The time going into the wind is 2 hours, giving us the first equation: \[ 180 = (v_p - v_w) \times 2 \]
- On the return, the trip takes 1.2 hours with the wind aiding, yielding the second equation: \[ 180 = (v_p + v_w) \times 1.2 \]
Simplifying these, we get:- \( 90 = v_p - v_w \)- \( 150 = v_p + v_w \)
By solving this system of equations, we can find unique values for \( v_p \) and \( v_w \). Adding the equations helps us eliminate \( v_w \), while substitution helps us solve for both variables. If you are new to systems of equations, practice identifying shared variables and setting equations accordingly.

Unit Conversion

Sometimes, operations with different units can complicate a problem, so it is crucial to handle **unit conversion** carefully. For this particular exercise, you're asked to convert time from minutes to hours. The return trip is described as taking 1 hour and 12 minutes.
- To convert minutes to hours, remember that 1 hour = 60 minutes.

• 12 minutes is equivalent to \( \frac{12}{60} = 0.2 \) hours.

- Thus, the total travel time becomes \( 1 + 0.2 = 1.2 \) hours.
Correct unit conversion ensures that all terms in the equations are consistently in the same unit, which is crucial for accurate calculations. Always ensure final values are converted to the desired unit for comprehensibility.

Distance-Rate-Time Problems

These are a central part of algebra word problems and involve the formula **Distance = Rate \( \times \) Time**. Understanding how to apply this formula is key to solving such problems. In this instance:

• The **distance** for each leg of the journey is the same: 180 miles.
• The **rate** is the effective speed of the airplane, which is either increased or decreased by the wind speed (\( v_w \)).
• The **time** varies between the journey to and from Bismarck due to the influence of the wind.

The main challenge is correctly setting the rate for each segment:- For the trip going into the wind, the effective speed is \( v_p - v_w \).- For the return trip, with wind assistance, it is \( v_p + v_w \).
By aligning these correctly with the formula, you ensure that your equations accurately reflect the scenario described in the problem. Practice will make these setups intuitive over time, helping you handle more complex word problems confidently.