document.write( "Question 1120070: A man flies a small airplane from Fargo to Bismarck, North Dakota a distance of 180 miles. Because he is flying into a headwind, 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 12 minutes. What is his speed in still air, and how fast is the wind blowing?
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Algebra.Com's Answer #735739 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "The solution can be obtained informally on a path nearly identical to the formal algebraic solution provided by the other tutor. \n" ); document.write( "The average speed against the wind is 180 miles / 2 hours = 90mph. The average speed with the wind is 180 miles / 1.2 hours = 150mph. \n" ); document.write( "One speed is the speed of the airplane plus the speed of the wind; the other is the speed of the airplane minus the speed of the wind. \n" ); document.write( "That is a kind of situation that shows up in many different kinds of problems. Logical analysis says that the speed of the airplane is halfway between the two average speeds; then the wind speed is the difference between the plane's speed and average speed on either of the legs. \n" ); document.write( "Halfway between 90 and 150 is 120. The speed of the airplane is 120mph; the speed of the wind is 150-120 = 120-90 = 30mph. \n" ); document.write( " |