Chapter 7.2, Problem 41E

To determine

The solution of airspeed of the plane and speed of the wind.

Answer to Problem 41E

Airspeed of the plane is $550\frac{miles}{hour}$

Speed of the wind is $50\frac{miles}{hour}$ ..

Explanation of Solution

Given information:

An airplane flying into a headwind travels the $1800-$ mile flying distance between Pittsburgh, Pennsylvania, and phoenix, Arizona, in $3$ hours and $36$ minutes. On the return flight, the airplane travels this distance in $3$ hours. Find the airspeed of the plane and speed of the wind, assuming that both remain constant.

Calculation:

Let speed of the plane $=x$

Let speed of the wind $=y$

According to the given information the first equation represents the airplane flying against the wind, so the rate is the speed of the plane minus the speed of the wind. The second equation represents the airplane flying with the wind, so the rate is the speed of the plane plus the speed of the wind as shown below:

  $\begin{array}{l}1800=(x-y)\left(3\frac{36}{60}\right)\\ 1800=(x+y)(3)\end{array}$

Simplify by converting mixed numbers into improper fractions:

  $\begin{array}{l}1800=(x-y)\left(\frac{216}{60}\right)\\ 1800=(x+y)(3)\end{array}$

Multiply the first equation by $\frac{60}{216}$ to isolate the variables. Multiply the second equation by $\frac{1}{3}$ to isolate the variables.

  $\begin{array}{l}\frac{60}{216}\left[1800=(x-y)\left(3\frac{36}{60}\right)\right]\text{ (1)}\\ \frac{1}{3}\left[1800=(x+y)(3)\right]\text{ (2)}\end{array}$

Distribute to simplify the expression:

  $\begin{array}{l}500=x-y\\ 600=x+y\end{array}$

Add the equations together to cancel out the $y$ variable:

  $1100=2x$

Solve for variable $x$ :

  $x=550$

Plug $x$ into the first equation and solve for $y$ :

  $\begin{array}{l}500=550-y\\ -50=-y\\ y=50\end{array}$

The airspeed of the plane is $550\frac{miles}{hour}$ , and the speed of the wind is $50\frac{miles}{hour}$ .