Question 653001
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Let the direction of travel for the initial leg of the trip be *[tex \LARGE X^\circ].  Then the direction of travel for the return leg is *[tex \LARGE \left(X^\circ\ +\ 180^\circ\right)\ \text{mod}\ 360^\circ].  The vector sum of the two opposing vectors is the difference of the magnitudes in the direction of the greater magnitude, i.e. 20 mph @ *[tex \LARGE \left(X^\circ\ +\ 180^\circ\right)\ \text{mod}\ 360^\circ].


The average speed is the total distance traveled divided by the total travel time.  The time for the outward leg is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{40}\ \ ]hours.


Where the one-way distance is *[tex \LARGE d]


The time for the return leg is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{60}\ \ ]hours.


The sum of these times is the total travel time:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{40}\ +\ \frac{d}{60}\ =\ \frac{100d}{2400}\ =\ \frac{d}{24}\ \ ]hours.


The total distance travelled is *[tex \LARGE 2d], so the average speed is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2d}{\frac{d}{24}}\ =\ 48\ \ ]mph.


If the trip took 5 hours...


If you go 48 mph for 5 hours, how far did you go?


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