Question

An airplane maintaining a constant airspeed takes as long to travel 260 miles with the wind as it does to travel 180 miles against the wind. If the wind is blowing at 20 mph, what is the rate of the plane in still air?

Answer 1

Given:

Distance travelled with the wind = 260 miles

Distance travelled against the wind = 180 miles

Speed of wind = 20 mph.

Let's find the rate of the plane in still air.

Apply the formula:

`rate=(distance)/(time)`

Thus, we have the equations:

`\begin{gathered} t_{with\text{ wind}}=(260)/(r+20) \\ \\ t_{against\text{ wind}}=(180)/(r-20) \end{gathered}`

Now, the rate in still air will be:

`\begin{gathered} t_{with\text{ wind}}=t_{against\text{ wind}} \\ \\ (260)/(r+20)=(180)/(r-20) \end{gathered}`

Now, let's solve for r in the equation above.

Cross multiply:

`180(r+20)=260(r-20)`

Divide both sides of the equation by 20:

`\begin{gathered} (180(r+20))/(20)=(260(r-20))/(20) \\ \\ 9(r+20)=13(r-20) \\ \\ \text{ Apply distributive property:} \\ 9r+9(20)=13r+13(-20) \\ \\ 9r+180=13r-260 \end{gathered}`

Solving further:

`\begin{gathered} 13r-9r=180+260 \\ \\ 4r=440 \end{gathered}`

Divide both sides by 4:

`\begin{gathered} (4x)/(4)=(440)/(4) \\ \\ x=110 \end{gathered}`

Therefore, the rate in still air is 110 mph.

ANSWER:

110 mph