Question 353603
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That's because your assertion that the average speed is 50 mph is incorrect.  The average speed is the total distance, 240 miles, divided by the total time, 5 hours, which is equal to 48 miles per hour.


Here's another illustration of the same thing.  You have a convenience store that is some distance *[tex \Large d] from your house.  When you drive to the store, there is some traffic, so you can only manage 25 miles per hour.  On the return trip you make 35 miles per hour.  What is your average speed?


It takes you *[tex \Large t_1] hours to make the outbound trip.  It takes *[tex \Large t_2] hours to come home.  Since you go 25 mph outbound we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_1\ =\ \frac{d}{25}]


Likewise


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_2\ =\ \frac{d}{35}]


So the total time is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{25}\ +\ \frac{d}{35}\ =\ \frac{60d}{875}]


The total distance traveled is *[tex \Large 2d] because you go *[tex \Large d] out and *[tex \Large d] back again, hence the average rate is then the total distance divided by the total time:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_{avg}\ =\ \frac{2d}{\frac{60d}{875}}\ =\ 2d\left(\frac{875}{60d}\right)\ =\ 29.1\overline{6}]


29 and 1/6 miles per hour, not 30 as most people would guess.


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