posted by jason on .
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?
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.