Question 524011
<font face="Times New Roman" size="+2">


It doesn't make any difference how far they travel given the time and speed relationships.


Distance:  *[tex \Large d] distance units


Rate(speed) of freight train:  *[tex \Large r] distance units per time unit


Rate of passenger train:  *[tex \Large r\ +\ 20] distance units per time unit


Time for freight to travel *[tex \Large d] distance units: *[tex \Large t] time units


Time for passenger train to travel *[tex \Large d] distance units: *[tex \Large \frac{t}{2}] time units


Time for freight train to go *[tex \Large d] distance units at a rate of *[tex \Large r] distance units per time unit:


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


Time for passenger train to go *[tex \Large d] distance units at a rate of *[tex \Large r\ +\ 20] distance units per time unit:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{t}{2}\ =\ \frac{d}{r\ +\ 20}]


Multiply the 2nd equation by 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{2d}{r\ +\ 20}]


Now that you have two different expressions equal to *[tex \Large t], set them equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2d}{r\ +\ 20}\ =\ \frac{d}{r}]


Cross-multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2dr\ =\ dr\ +\ 20d]


Multiply both sides by *[tex \Large \frac{1}{d}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2r\ =\ r\ +\ 20]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ 20]


So the freight averages 20 distance units per time unit, and the passenger train goes 40 distance units per time unit.  Since you were working with miles as  distance unit and hours as time units, your answers should be expressed in miles per hour.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>