Question 605313
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I'm not so sure that I would hang my hat on your algebra techniques, but you did get the right answer.  I don't quite understand why you thought you needed to set this up as an absolute value problem unless Peoria was an intermediate stop and you wanted to know the <i>two different times</i> that she would be 80 miles distant from Peoria (The first time being when she had 80 miles remaining of the 180 mile trip and the second being 1 hour and 36 minutes after she had passed Peoria).


Assuming, as one would ordinarily do when told that Person 1 is driving from point A to point B, that the person stops when they get to point B, then you have no reason to use the absolute value function since, given that she stops when she gets to Peoria, the largest value that the quantity 50t can assume is 180 miles, and therefore the expression 180 - 50t can never be less than zero.


With that assumption in mind, if she only has 80 miles to go, then she has traveled 100 miles and 100 miles divided by 50 mph is 2 hours.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 180\ -\ 50t\ =\ 80]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -50t\ =\ 80\ -\ 180]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -50t\ =\ -100]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ 2].


If you actually were interested in the next time that she would be 80 miles from Peoria if she had continued on through, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |180\ -\ 50t|\ =\ 80]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 180\ -\ 50t\ =\ 80]


which we solved above, or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 180\ -\ 50t\ =\ -80]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -50t\ =\ -80\ -\ 180]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -50t\ =\ -260]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ 5.2] hours


or 5 hours and 12 minutes after leaving Chicago, which I will leave as an exercise for the student to verify is 1 hour and 36 minutes after passing through Peoria.


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