One way G can catch S is as follows (it is not the fastest way).

G waits until he knows that S has traveled for one mile. At that time, both S and G are somewhere on a circle with radius one mile, and with its center at the original position of S. G then begins to travel with a velocity that has a radially outward component equal to that of S, and with a tangential component as large as possible, given G's own limitation of total speed. By doing so, G and S will always both be on an identical circle having its center at the original position of S. Because G has a tangential component whereas S does not, G will always catch S (actually, this is not proven until you solve the o.d.e. associated with the problem).

If G can go at 40 mph and S goes at 20 mph, you can work out that it will take G at most 1h 49m 52s to catch S. On average, G will catch S in:

( -2pi + sqrt(3) ( exp(2pi/sqrt(3)) - 1 )) / 40pi hours,

which is, 27 min and 17 sec.

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