Question 596745
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At 10:30 when James starts, John will be 30 minutes into his trip.  30 minutes at 2 km/hr is 1 km.  One hour later John will be 3 km from the start and James will be 4 km from the start, making a 3-4-5 right triangle at time 11:30.


That was the easy way.  Now the hard way:


In the first 30 minutes John goes 1 km.  Every hour after that, he goes 2 km.  So counting *[tex \LARGE t] hours since JAMES starts, John goes *[tex \LARGE 2t\ +\ 1] km.  James is going twice as fast as John, so every hour after James starts he goes *[tex \LARGE 4t] km.


In order for them to be 5 km apart:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2t\ +\ 1)^2\ +\ (4t)^2\ =\ 25]


Simplifying:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5t^2\ +\ t\ -\ 6\ =\ 0]


Solve for the *[tex \LARGE t].  Toss out the negative value (you don't care what happened <i>before</i> either of them started the trip, do you?)


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