Question 268137
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Remember, distance equals rate times time.


Let *[tex \Large d] represent the distance traveled by the slower car at *[tex \Large r\ =\ 45] mph for some amount of time, *[tex \Large t].


Then *[tex \Large d\ +\ 60] is the distance traveled by the faster car at *[tex \Large r\ =\ 60] mph for the same amount of time, *[tex \Large t]


The trip taken by the slower car can be described as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 45t]


And the trip taken by the faster car can be described as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ +\ 60\ =\ 60t]


which can be arranged thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 60t\ -\ 60]


Now we have two expressions for *[tex \Large d] which we can set equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 60t\ -\ 60\ =\ 45t]


Solve for *[tex \Large t] for the answer to your problem.


<b><i>Super Double Plus Extra Credit:</i></b> What was the distance traveled by each of the cars in the amount of time you computed above?  Do the distances differ by 60 miles?


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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