Problem 27

Question

Dave and Sandy Hartranft are frequent flyers with Delta Airlines. They often fly from Philadelphia to Chicago, a distance of 780 miles. On one particular trip they fly into the wind, and the flight takes 2 hours. The return trip, with the wind behind them, takes only $1 \frac{1}{2}$ hours. Find the speed of the wind and find the speed of the plane in still air.

Step-by-Step Solution

Verified

Answer

The speed of the plane is 455 mph, and the wind speed is 65 mph.

1Step 1: Define Variables

Let $ p $ represent the speed of the plane in still air and $ w $ represent the speed of the wind. The distance of the trip is 780 miles.

2Step 2: Set Up Equations for Each Flight

For the flight into the wind, the effective speed is $ p - w $ and it takes 2 hours to cover 780 miles. The equation is:\[ 780 = 2(p - w) \]For the return trip with the wind, the effective speed is $ p + w $ and it takes 1.5 hours to cover 780 miles. The equation is:\[ 780 = 1.5(p + w) \]

3Step 3: Solve for Plane Speed Equation

Rearrange both equations:\[ p - w = \frac{780}{2} = 390 \]\[ p + w = \frac{780}{1.5} = 520 \]

4Step 4: Add and Solve the System of Equations

Add the two equations to eliminate $ w $:\[(p - w) + (p + w) = 390 + 520 \2p = 910 \p = \frac{910}{2} = 455 \]

5Step 5: Solve for Wind Speed with Plane Speed Known

Substitute $ p = 455 $ into one of the original equations (e.g., $ p - w = 390 $ ) :\[455 - w = 390 \w = 455 - 390 = 65\]

Key Concepts

Systems of EquationsDistance-Rate-Time ProblemsLinear Equations

Systems of Equations

In algebra, a system of equations is a collection of two or more equations that you deal with all together. This approaches allow us to find the values of unknown variables which satisfy all those equations. Here's how it works in the problem:

We defined two variables: let $ p $ be the speed of the plane in still air, and $ w $ be the speed of the wind.
The two crucial equations derived from the problem are:
- For flying into the wind: $ 2(p - w) = 780 $
- For flying with the wind: $ 1.5(p + w) = 780 $

To solve this system of equations, you can use either substitution or elimination methods. Here, we use the elimination method since it involves fewer steps.
First, rearrange the equations to isolate $ p - w $ and $ p + w $. Then, add the two equations together:

$ p - w = 390 $
$ p + w = 520 $

Adding these, the $ w $ terms cancel out, simplifying our solution process. This gives us the speed of the plane $ p $, which we will explore further in the context of the actual problem.

Distance-Rate-Time Problems

Distance-rate-time problems are a key component of algebra, particularly when dealing with motion. These problems involve understanding how three main factors—distance, rate (speed), and time—relate to each other through the equation:
\[ \text{Distance} = \text{Rate} \times \text{Time} \]For Dave and Sandy’s flights, we use this formula to relate how fast they're traveling over a given distance within a certain timeframe.

For the flight into the wind, we see negative wind effects, reducing the plane's effective speed.
On the return trip with the wind, the wind boosts their speed.

The calculations break down like this:

Flight against the wind: $ 780 = 2(p-w) $, which simplifies to an effective slower speed $ p - w $.
Flight with the wind: $ 780 = 1.5(p+w) $, illustrating the wind's assistance in reducing travel time.

Applying the distance-rate-time concept here allows us to investigate different factors that affect flying times, leading to finding both the wind's speed and the plane's speed in still air.

Linear Equations

Linear equations are foundational in algebra. They involve two variables and form a straight line when graphed on a coordinate plane. The equations in this exercise are linear and appear like this: