<PRE><B>Question 1125359</B>
 A jet flying at an altitude of 35,000 ft passes over a small plane flying at 10,000 ft headed in the same direction.
 The jet is flying twice as fast as the small plane, and 30 mins later they are 100 mi apart.
 Find the speed of each plane.
:
Their difference in flight paths are 25,000 ft which is 4.7 mi apart but that is not enough to make a difference in the way we solve this problem.
:
let s = the speed of the slower plane
then
2s = the speed of the jet
:
Write a distance equation, dist = time * speed
{{{1/2}}}*(2s-s) = 100
2s - s = 200
s = 200 mph is speed of the slower plane
then obviously
400 mph is the speed of the jet
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