Math
posted by David .
an air traffic controller spots two planes at the same altitude flying towards one another. their flight paths form a right angle at point p. One plane is 150 miles from point p and is moving 450mph. the other plane is moving at 450mph but is 200 miles from point p. write the distance d between the planes as a function of time t.

the planes' distances from p form a scaledup 345 right triangle, so at the time specified, d, the hypotenuse, is 250
at time t hours later,
d^2 = (150450t)^2 + (200450t)^2
That's kind of nasty, so a simpler formula would be,
Let x be the distance of the first plane. Then the second plane is 4/3 x away from p.
d^2 = x^2 + (4/3 x)^2 = 25/9 x^2
since x = 150250t,
d^2 = 25/9 (150450t)^2
= 25/9 * 150^2 (13t)^2
= 25*2500 (13t)^2
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