Question 1189258
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Let *[tex \Large t] represent the total time of the trip and *[tex \Large d] represent the total distance traveled.  The average speed is equal to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{t}]


There are three parts of this trip, so let the three distances be represented by *[tex \Large d_1\,],*[tex \Large d_2\,], and *[tex \Large d_3] and we know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  d\ =\ d_1\ +\ d_2\ +\ d_3]


Also, if we have accounted for one-third of the trip and one-fourth of the trip, the rest of the trip must amount to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \frac{t}{3}\ -\ \frac{t}{4}\ =\ \frac{5t}{12}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_1\ =\ \(20\,\text{km/hr}\,\cdot\,\frac{t} {3}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_2\ =\ \(30\,\text{km/hr}\,\cdot\,\frac{t} {4}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_3\ =\ \(50\,\text{km/hr}\,\cdot\,\frac{5t}{12}\)]


So

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \frac{20t}{3}\ +\ \frac{30t}{4}\ +\ \frac{250t}{12}]


But the average speed is the total distance divided by the total time, *[tex \Large t], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s_{avg}\ =\ \frac{20}{3}\ +\ \frac{30}{4}\ +\ \frac{250}{12}]


You can do your own arithmetic.

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

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