Question 179770
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The ground speed of the aircraft for the trip to Europe is given by *[tex \Large r = {4800 \over 5} = 960] kilometers per hour.  The ground speed for the return trip is given by *[tex \Large r = {4800 \over 6} = 800] kmph.


When going in the same direction as the wind, i.e. the trip to Europe, the wind speed adds to the aircraft's air speed to form the ground speed.  On the return, the wind speed subtracts from the air speed to form the ground speed.


This relationship gives us the following system of linear equations:


*[tex \Large r_a + r_w = 960]


*[tex \Large r_a - r_w = 800]


Since the coefficients on *[tex \Large r_w ] in the two equations are additive inverses, add the two equations term-by-term:


*[tex \Large 2r_a = 1760]


Solve:


*[tex \Large r_a = 880]


Hence, the aircraft air speed is 880 kmph, and the wind speed must be *[tex \Large r_w = 960 - r_a = 960 - 880 = 80] kmph



<i><b>Super-Double-Plus Extra Credit:</b></i>  From the information given in the problem can you determine the wind <i><b>velocity</b></i>?  If not, what additional data element would you require?



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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