SOLUTION: On a flight from Chicago to Los Angeles, there was a tailwind (with wind). On the return flight that same day, the airplane was flying into a headwind (against the wind). Suppose

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Question 957321: On a flight from Chicago to Los Angeles, there was a tailwind (with wind). On the return flight that same day, the airplane was flying into a headwind (against the wind). Suppose an airplane has an average speed of 450 mph in still air and the wind was a constant speed for both flights, what was the speed of the wind if Chicago and Los Angeles are 2015 miles apart and the flights took a combined time of 9 hours.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
W=wind speed
2015mi%2F%28450mph%2BW%29%2B2015mi%2F%28450mph-W%29=9hrs Multiply by (450mph-W)(450mph+W)
2015mi%28450mph-W%29%2B2015%28450mph%2BW%29=9hrs%28450mph-W%29%28450mph%2BW%29

1813500%28mi%5E2%2Fhr%29=1822500%28mi%5E2%2Fhr%5E2%29-9W%5E2 Add 9W^2-1813500mi^2hr^2 to each side.
9W%5E2=%289000%29%28mi%5E2%2Fhr%5E2%29 Divide each side by 9.
W%5E2=%281000%29%28mi%5E2%2Fhr%5E2%29Find square root of each side.
W=31.62mph ANSWER: The wind speed was 31.62 miles per hour.
CHECK:
2015mi%2F%28450mph-31.62mph%29%2B2015mi%2F%28450mph%2B31.62mph%29=9hrs
2015mi%2F418.38mph%2B2015mi%2F481.62mph=9hrs
4.82hrs%2B4.18hrs=9hrs
9hrs=9hrs