SOLUTION: If a car is traveling 83 mph and passes a stopped police car that then gives pursuit and the police car catches the car traveling 83 mph in one mile. How fast is the police car tra

Question 827185: If a car is traveling 83 mph and passes a stopped police car that then gives pursuit and the police car catches the car traveling 83 mph in one mile. How fast is the police car traveling? also please compute for 1.5 miles also.
Thanks - my 20 year old son has court tomorrow morning & this just does not make sense to me!
Found 2 solutions by Alan3354, Fermat:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
If a car is traveling 83 mph and passes a stopped police car that then gives pursuit and the police car catches the car traveling 83 mph in one mile. How fast is the police car traveling? also please compute for 1.5 miles also.
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There's info missing.
How long after being passed did the cop car start to move?
What was the acceleration of the cop car?
You can't get an accurate solution without knowing that.
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I agree it doesn't sound right.
I can make some estimations later.
Answer by Fermat(136) (Show Source): You can put this solution on YOUR website!
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.

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