Question 827185
There is no information about the police car's movement, what its top speed is or what its initial acceleration might be.
I did some research. and found the fastest police car in the US is a Cadillac CTS-V, which has a top speed of 163 mph and can do 0 -> 60 mph in 3.9 secs.

Son's car:
Travelling at 83 mph over 1 mile.
Therefore journey time = (1/83) hr = (60/83) min = (3600/83) sec = 43 secs.
So any police car would have to accelerate, (up to top speed say) and then catch your son within 1 mile.

Police car:
Top speed: 163 mph (239 ft/sec)
Acceleration: 0 -> 60 mph in 3.9 secs, i.e. accln = a = 88/3.9 = 22.54 ft/sec^2
Assume same accln to reach top speed.
Time to reach top speed = t1 = v_max/a = 239/22.54 = 11 secs, say.
Distance reached by police car in 1st 11 secs is,
s1 = (1/2)*a*t^2 = (1/2)*22.544*11^2 = 1365 ft = 455 yds
Distance covered by your son in 11 secs is (83/60)*88*11 ft = 1339 ft = 446 yds.
So police will have already caught up to your son before they have finished accelerating, in just over 1/4 mile.

2nd scenario ( a slower police car)
Top speed: 150 mph (220 ft/sec)
Acceleration: accln = a = 20 ft/sec^2, say.
Assume this accln to reach top speed.
Time to reach top speed = t1 = v_max/a = 220/20 = 11 secs.
Distance reached by police car in 1st 11 secs is,
s1 = (1/2)*a*t^2 = (1/2)*20*11^2 = 1210 ft = 403 yds
Distance covered by your son in 11 secs is (83/60)*88*11 ft = 1339 ft = 446 yds.
Your son is 446 - 403 = 43 yds ahead of the police car.
Police car's speed is v = 220 ft/s
Son's speed = 83 mph = (83/60)*88 ft/sec = 121.7 ft/sec
Speed difference = 220 - 121.7 = 98.3 ft/sec
Time to catch son = (dist ahead)/(speed diff) = (43*3)/(98.3) = 1.3 sec
Distance travelled by son = 446*3 + 1.3*121.7 = 1496 ft = 499 yds, much less than (1/2) a mile.

3rd scenario ( a much slower police car)
Top speed: 120 mph (176 ft/sec)
Acceleration: accln = a = 15 ft/sec^2, say.
Assume this accln to reach top speed.
Time to reach top speed = t1 = v_max/a = 176/15 = 12 secs.
Distance reached by police car in 1st 12 secs is,
s1 = (1/2)*a*t^2 = (1/2)*15*12^2 = 1080 ft = 360 yds
Distance covered by your son in 12 secs is (83/60)*88*12 ft = 1461 ft = 487 yds.
Your son is 487 - 360 = 127 yds ahead of the police car.
Police car's speed is v = 176 ft/s
Son's speed = 121.7 ft.sec
Speed difference = 176 - 121.7 = 54.3 ft/sec
Time to catch son = (dist ahead)/(speed diff) = (127*3)/(54.3) = 7 sec
Distance travelled by son = 487*3 + 7*121.7 = 2313 ft = 771 yds, just under 1/2 mile.

If you continued in the same vein, you would find that you needed a police car with a top speed of 100 mph and an accln of 10 ft/sec^2 to catch your son after a chase of 1 mile.

4th scenario ( a very slow police car)
Top speed: 100 mph (146.67 ft/sec)
Acceleration: accln = a = 10 ft/sec^2, say.
Assume this accln to reach top speed.
Time to reach top speed = t1 = v_max/a = 146.67/10 = 15 secs.
Distance reached by police car in 1st 15 secs is,
s1 = (1/2)*a*t^2 = (1/2)*10*15^2 = 1125 ft = 375 yds
Distance covered by your son in 15 secs is (83/60)*88*15 ft = 1826 ft = 608 yds.
You son is 608 - 375 = 233 yds ahead of the police car.
Police car's speed is v = 146.67 ft/s
Son's speed = 121.7 ft.sec
Speed difference = 146.67 - 121.7 = 25 ft/sec
Time to catch son = (dist ahead)/(speed diff) = (233*3)/(25) = 28 sec
Distance travelled by son = 608*3 + 28*121.7 = 5232 ft = 1748 yds, just under a mile.
OK, the above takes a lot of reading, but I've tried to cover a lot of area to try and assure you that it is perfectly possible for a police car to catch a speeding car, from a standing start, within 1 mile.