Question 575458
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At *[tex \Large t\ =\ 0], A is 40 miles from O and B is at O.  As time increases, A moves away from O at 40 mph, so at any positive time *[tex \Large t] hours, A is *[tex \Large 40\ +\ 40t] miles from O.  At the same time, B is *[tex \Large 60t] miles from O, but in a direction at right angle to A's direction of travel.


Hence, the distance A has traveled at time *[tex \Large t] is one leg of a right triangle and the distance B has traveled at the same time is the other leg of a right triangle.


With a tip of the hat to Mr. Pythagoras:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d(t)\ =\ \left((60t)^2\ +\ (40\ +\ 40t)^2\right)^{\frac{1}{2}}]


You can simplify if you like.


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