Question 284519
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Flying into the headwind, the ground speed is airspeed minus wind speed, so 150 - 30 is 120.  Flying with a tail wind, the ground speed is airspeed plus wind speed, here: 180 mph.


Distance equals rate times time.  Let *[tex \Large t] represent the time flying into the wind.  Then, since the total trip has to be no more than 20 hours, the time flying home must be *[tex \Large 24\ -\ t].  And the distance, *[tex \Large d], out has to be equal to the distance coming back.


So:


Outward trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 120t]


Return trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 180(20\ -\ t)]


But since *[tex \Large d\ =\ d]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 120t\ =\ 180(20\ -\ t)]


Now solve for *[tex \Large t]


Once you have a value for *[tex \Large t], substitute it into either of the two equations and solve for d.



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