Question 597320
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If the guy who starts at 10am is going 50 mph, he will have gone 50 miles when the other guy starts.  Which is to say that when the second guy starts they will be 170 minus 50 = 120 miles apart.


Distance equals rate times time, so time equals distance divided by rate.  The distance the 50mph guy will cover of the remaining 120 miles can be represented by *[tex \LARGE x], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{x}{50}]


The other guy must be traveling what remains of the 120 after you subtract the *[tex \LARGE x] miles the first guy will cover, so his trip is described by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{120\ -\ x}{30}]


Since we are concerned with the time they meet, the time each of them is on the road (starting at the 11:00 am time) is equal, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{50}\ =\ \frac{120\ -\ x}{30}]


Solve for *[tex \LARGE x]


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