Question 202309
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Use the relationship between distance, rate, and time:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d = rt]


Which can be expressed as time as a function of distance and rate as:


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


Let *[tex \Large d] represent the distance John walked, and then *[tex \Large 448 - x] represents the distance that Peter walked. So, using the given rates we can say, with regard to John:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{d}{4}]


And with regard to Peter:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{448-d}{3}]


Since they started at the same time, and the end of both trips is the time that they met, *[tex \Large t] is identical for both equations.  That means we can equate the right-hand sides of the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{4} = \frac{448-d}{3}]


Solve this equation for *[tex \Large d].  Hint: Cross-multiply, distribute, collect like terms, and then divide by the resulting coefficient on d.


*[tex \Large d] will be the distance that John walked.  Subtract this value from 448 to get the distance that Peter walked.  That will answer question 2.


Divide the distance John walked by 4 or divide the distance Peter walked by 3 (they are both the same result) to get the number of seconds that will have elapsed when they met.  Divide this result by 60 to get the number of minutes for question 1.


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