Question 1078220
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Begin with *[tex \Large d\ =\ rt], <i>distance equals rate times time</i>


As for the initial trip, we do not know the distance and we do not know the time taken.  So all we know is that *[tex \Large d\ =\ 60t], or more conveniently, *[tex \Large 60\ =\ \frac{d}{t}].


As for the return trip, we have a similar relationship, namely *[tex \Large 40\ =\ \frac{d}{t_1}].  But since *[tex \Large 40\ =\ \frac{2}{3}\ \cdot\ 60], we know that the return trip must have taken *[tex \Large \frac{3}{2}\ =\ 1.5] times as long as the initial trip.  Hence, *[tex \Large t_1\ =\ 1.5t].


The average speed is then the total distance, *[tex \Large 2d], divided by the total time, *[tex \Large t\ +\ 1.5t\ =\ 2.5t], which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S_{avg}\ =\ \frac{2d}{2.5t}]


However, we know from the relation describing the initial trip that *[tex \Large \frac{d}{t}\ =\ 60], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  S_{avg}\ =\ \frac{2}{2.5}\ \cdot\ 60]


You can do your own arithmetic.


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