Chapter 3.6, Problem 8P

To determine

To find:the plans air speed and the wind speed.

Answer to Problem 8P

The plans air speed and the wind speed are 275 miles per hour and 25 miles per hour, respectively.

Explanation of Solution

Given information:

With a head wind, a plane traveled 1000 mi in 4 hours.

With a tail wind, a plane traveled 1000 mi in 3 hours 20 min.

Calculation:

Let x be the air speed of the plane and let y be speed of wind.

With a tail wind, the ground speed is $x+y$ .

With headwind, the ground speed is $x-y$ .

Note that, the distance traveled by plane is multiplication of ground of speed and time.

With tail wind, the plane traveled 1000 mi in 3 hours20 min.

  $\begin{array}{l}1000=\left(x+y\right)\times 3\text{ hr 20 min}\\ 1000=\left(x+y\right)\left(3+\frac{1}{3}\right)\left[\because \text{ 3 hr 20 min}=3+\frac{20}{60}\right]\\ 1000=\left(x+y\right)\left(\frac{10}{3}\right)\\ 3000=10x+10y\left[\text{By cross multiplication}\right]\end{array}$

  $10x+10y=3000$   ....... (1)

With head wind, the plane traveled 1000 mi in 3 hours 20 min.

  $\begin{array}{l}1000=\left(x-y\right)\times 4\\ \frac{1000}{4}=x-y\end{array}$

  $x-y=250$   ....... (2)

Compute the solutions of the system $x-y=250$ and $10x+10y=3000$ by linear combination method.

Multiply the equation $x-y=250$ by 10. That is, $10x-10y=2500$

Add two equations $10x-10y=2500$ and $10x+10y=3000$ , simplify further.

  $\begin{array}{l}10x-10y=2500\\ 10x+10y=3000\\ \overline{20x+\text{ 0}y=5500}\end{array}$

Divide on both sides by20,

  $\begin{array}{l}20x=5500\\ x=\frac{5500}{20}\\ x=275\end{array}$

Substitute $x=275$ in $x-y=250$ and simplify the value of y ,

  $\begin{array}{l}275-y=250\\ -y=250-275\\ -y=-25\\ y=25\end{array}$

Therefore, the plans air speed and the wind speed are 275 miles per hour and 25 miles per hour, respectively.