A man flies a small airplane from Fargo to Bismarck, North Dakota \( -\cdots \) a distance of 180 miles. 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 hour and 0 minutes. What is the plane's speed in still air, and how fast is the wind blowing? his plane speed equals I the wind speed equals

We are given that the airplane travels 180 miles from Fargo to Bismarck with a headwind, taking 2 hours, and on the return trip with a tailwind it takes 1 hour. Let \( p \) be the speed of the plane in still air (in mph) and \( w \) be the wind speed (in mph). **Step 1. Set up the equations.** For the trip with the headwind, the effective speed is \( p - w \) and the time taken is 2 hours: \[ p - w = \frac{180}{2} = 90. \] For the return trip with the tailwind, the effective speed is \( p + w \) and the time taken is 1 hour: \[ p + w = \frac{180}{1} = 180. \] **Step 2. Solve the system of equations.** We have: \[ \begin{aligned} p - w &= 90, \\ p + w &= 180. \end{aligned} \] Add the two equations: \[ (p - w) + (p + w) = 90 + 180, \] which simplifies to: \[ 2p = 270. \] So the speed of the plane in still air is: \[ p = \frac{270}{2} = 135. \] Substitute \( p = 135 \) into the first equation: \[ 135 - w = 90. \] Solving for \( w \): \[ w = 135 - 90 = 45. \] **Step 3. Final answers.** The plane's speed in still air is \( 135 \) mph, and the wind speed is \( 45 \) mph.