Posted by jason on Monday, March 16, 2009 at 2:32am.
Let's measure position from the ground location at noon, where the train was when the plane was overhead. It will make the math simpler but will not affect the answer. North position from there will have a "j" unit vector and west poeitiom from there will have a "-i" unit vector. The vertical unit vector is k.
At the time in question, the airplane position vector is
R2 = (5 + 500 t) j + 4 k
and the train position vector is
R1 = -(1 + 80 t)i
The separation vector is
D = R2 - R1 = (1 + 80 t)i + (5 + 500 t) j + 4 k
The rate of change of this vector is
dD/dt = 80 i + 500j mph
Take the square root of the sum of the squares for the rate the distance in increasing.
That would be 506.4 mph
Note that the altitude and the separation at that time do not matter, only the speeds.
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