Question 1116842
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Basic formula is *[tex \Large d\ =\ rt]


For the outbound trip (with the wind), the speed of the aircraft is *[tex \Large 125\ +\ r_w] where *[tex \Large r_w] is the rate of speed of the wind.


For the return trip (against the wind), the speed of the aircraft is *[tex \Large 125\ -\ r_w]


Let the time of the outbound trip be *[tex \Large t], then the time of the return trip must be whatever remains of the total trip time of 5 hours, or *[tex \Large 5\ -\ t].


Since the two legs of the trip are equal in distance, namely 300 miles, we have two equations in *[tex \Large t] and *[tex \Large r_w]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  300\ =\ (125\ +\ r_w)t]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  300\ =\ (125\ -\ r_w)(5\ -\ t)]


Although we are asked for the speed of the wind, it will be a much neater calculation to solve for *[tex \Large t] first.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 125\ +\ r_w\ =\ \frac{300}{t}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 125\ -\ r_w\ =\ \frac{300}{5\ -\ t}]


Add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 250\ =\ \frac{300}{t}\ +\ \frac{300}{5\ -\ t}]


Combine the RHS fractions over the common denominator *[tex \Large 5t\ -\ t^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 250\ =\ \frac{1500}{5t\ -\ t^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ =\ \frac{6}{5t\ -\ t^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2\ -\ 5t\ +\ 6\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (t\ -\ 2)(t\ -\ 3)\ =\ 0]


So *[tex \Large t] is either 2 or 3.  However, the way we set up the problem with *[tex \Large t] representing the time of the 'with the wind' leg of the trip, *[tex \Large t] must be the smaller of the two values.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ 2]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\ -\ t\ =\ 3]


So, since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 300\ =\ (125\ +\ r_w)*2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 150\ =\ 125\ +\ r_w]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_w\ =\ 25]


Checking the answer in the return trip equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 300\ =\ (125\ -\ r_w)*3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 100\ =\ 125\ -\ r_w]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_w\ =\ 25]


Checks


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