SOLUTION: An airplane flying north at 320 mph passed over a point on the ground at 2:00 p.m. Another plane at the same altitude passed over the point at 2:15 p.m.
flying east at 300 mph. A
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Question 657373: An airplane flying north at 320 mph passed over a point on the ground at 2:00 p.m. Another plane at the same altitude passed over the point at 2:15 p.m.
flying east at 300 mph. At what time will the planes be 500 miles apart?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
15 minutes after the northbound plane passes the point on the ground, it will be 0.25 times 320 = 80 miles north of the point on the ground. Let 
represent the number of hours after 2:15 that the two planes are 500 miles apart. The first plane will travel 
miles north of the 80 mile point north of the original flyover point, so the total distance traveled by the northbound plane starting at the flyover point will be 
. The total distance traveled by the eastbound plane will be 
.
These two distances are legs of a right triangle. We need to find such that the hypotenuse of the triangle measures 500 miles.
Using Pythagoras:
^2\ +\ (300t)^2\ =\ 500^2)
A little Algebra Music, Sammy...
Which factors:
(1924t\ +\ 2436)\ =\ 0)
Toss out the negative root. 
2:15 PM plus 1 hour is 3:15 PM.
John
My calculator said it, I believe it, that settles it
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