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October 1, 2014

October 1, 2014

Posted by **David** on Sunday, September 9, 2012 at 11:53am.

- Math -
**Steve**, Sunday, September 9, 2012 at 2:19pmthe planes' distances from p form a scaled-up 3-4-5 right triangle, so at the time specified, d, the hypotenuse, is 250

at time t hours later,

d^2 = (150-450t)^2 + (200-450t)^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 = 150-250t,

d^2 = 25/9 (150-450t)^2

= 25/9 * 150^2 (1-3t)^2

= 25*2500 (1-3t)^2

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