Question 421998
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Let *[tex \Large r] represent the air speed of the aircraft and let *[tex \Large r_w] represent the wind speed.  When the aircraft is flying in the same direction as the wind, the ground speed is *[tex \Large r\ +\ r_w].  When the aircraft is flying against the wind the ground speed is *[tex \Large r\ -\ r_w].


The average ground speed is the distance traveled divided by the time.  Hence the average ground speed while flying with the wind is *[tex \Large \frac{1280}{8}\ =\ 160].  The average ground speed flying against the wind is *[tex \Large \frac{1280}{12.8}\ =\ 100]  (Note: 6 minutes is 1/10 of an hour, so 48 minutes is 8/10s of an hour)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ +\ r_w\ =\ 160]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ -\ r_w\ =\ 100]


Multiply the 2nd equation by -1 and then add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0r\ +\ 2r_w\ =\ 60]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_w\ =\ 30]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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