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As long as both trains are moving they will always meet at a point 1/3 of the way between their two positions. Lets say the initial positions for the slow train is 0 and the fast train is 3x. Thus if they both left on time they would meet at point x. Let's assume that in 5 minutes the slow train can travel u units of distance and the fast train can cover 2u units.
Assume first the slow train leaves late. At the moment the slow train does leave the fast train will be at point 3x-2u. The meeting point will be 1/3 of the way between 0 and 3x-2u = (3x-2u)/3 = x-2u/3. We also know the distance from this point to x is 2 miles. In other words x-(x-2u/3)=2 ... 2u/3=2... u=3.
It is not necessary but we can also calculate u if the fast train leaves late. In this case when the fast train does leave from point 3x the slow train will already be at point u. The meeting point will be u+(3x-u)/3 = x+2u/3. Again the distance between this point and x is 2 miles. In other words (x+2u/3)-x=2 ... 2u/3=2 ... u=3.
Now we just plug what we know into the d=rt formula (distance=rate * time). The distance is one unit, which equals 3 miles. The time is 5 minutes, or 1/12 of an hour. So we have 3=r*(1/12) ... r=36 m.p.h..
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