Posted by **jason** on Monday, March 16, 2009 at 2:32am.

26. A train is heading due west from St. Louis. At noon, a

plane flying horizontally due north at a fixed altitude of

4 miles passes directly over the train. When the train has

traveled another mile, it is going 80 mph, and the plane

has traveled another 5 miles and is going 500 mph. At

that moment, how fast is the distance between the train

and the plane increasing?

- calculus -
**drwls**, Monday, March 16, 2009 at 5:06am
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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