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Assume a car is driving 73 MPH and passes a cop car at rest. Exactly 2 miles later, the cop catches up to the speeding car.

How fast did the cop travel to catch that car in 2 miles and what was his acceleration?


I'm stumped and it's driving me freaking nuts.

Fast enough to catch you and his acceleration was as enough as his foot would allow

I think the cop would have to have an acceleration of ~88.8 miles/hour.

The car travels the 2 miles in ~1.643836 minutes (2 / 73).

Cop car's acceleration:
d = vt + (1/2)at^2
d = 2 miles
v = initial velocity = 0
t = 1.643836 minutes

 Goremaul wrote:
I think the cop would have to have an acceleration of ~88.8 miles/hour.

That's assuming the flux capacitors are charged, and we're getting a full 1.21 Gigawatts output from them

 Goremaul wrote:
I think the cop would have to have an acceleration of ~88.8 miles/hour.

The car travels the 2 miles in ~1.643836 minutes (2 / 73).

Cop car's acceleration:
d = vt + (1/2)at^2
d = 2 miles
v = initial velocity = 0
t = 1.643836 minutes

If his acceleration was 88 mph, that would mean his top speed would be 144 mph--which doesn't seem right. This is using the acceleration formula;

A = (V2-V1)/Time

88 mph = (V2-0) / 1.64 minutes
88 * 1.64 = V2
V2 = 144

Or am I missing the conversion of mph to miles per minute?

Nah, that sounds about right. He'd have to be going pretty damn fast to catch up that quickly from rest.

The other thing you could try would be plugging it into a graph; might give you a good visual of where the cars are at any given point.

 Goremaul wrote:
Nah, that sounds about right. He'd have to be going pretty damn fast to catch up that quickly from rest.

The other thing you could try would be plugging it into a graph; might give you a good visual of where the cars are at any given point.

Yep, I did this a couple of different ways and that's indeed the answer. Thanks for the help Goremaul.

Ryan

No worries! Glad to be of help.

So...do police generally get up to 144mph when tracking down 'speeders' going 73mph?!?

Probably not because acceleration is not constant the whole time.

Well, 144mph is what they'd reach in order to travel the 2km in the same time that the other car does. In reality, he might reach 120 or so, and the other car might start to slow down to stop. That makes the math a bit more complicated though.

Or he might get to 100mph and take longer / further to catch up.

Now give us a more complicated problem that includes space-time, relativity, Lorentz! Lol.

 Alpharius wrote:
So...do police generally get up to 144mph when tracking down 'speeders' going 73mph?!?

Only in Ideal Physics World when driving spherical cars on a frictionless road.

My mistake - I thought AoE was giving us a 'real world example' but thanks for fulfilling my USRDA of "Internet Snark"!

Are we using the 5 or 10 phase movement system, and what’s the HC of the police cruiser?

/Car Wars

What if there was a second cop car further down the road like a "speed trap"

 Alpharius wrote:
My mistake - I thought AoE was giving us a 'real world example' but thanks for fulfilling my USRDA of "Internet Snark"!

Is that a thing? I've got some posts to make!

 Alpharius wrote:
My mistake - I thought AoE was giving us a 'real world example' but thanks for fulfilling my USRDA of "Internet Snark"!

It was indeed a real world example.

A physician I work with got the ticket--not me--but we were both trying to figure out the physics behind it (He was speeding and just laughed off the ticket but thought the cop's report was rather bogus).

Ok, if you want a real answer instead of a problem in Ideal Physics World (spherical cars, frictionless roads, etc) then looking at the constant-acceleration formula isn't very helpful. The cop isn't accelerating continuously until they catch the speeder (and immediately overshoot because they didn't slow down to match speeds), they're quickly accelerating to a decent speed and then closing in over a relatively long distance with a modest speed advantage. So let's say the cop has good reaction time, catches the speeder on radar before they reach the cop, and pulls onto the road to start the chase exactly as the speeder passes by at 73mph. A quick google search tells me that 0-60 times for a cop car are about 8 seconds, so round that up to about 10 seconds to match the speeder's 73mph. That means when the speeder has gone about a fifth of a mile down the road the cop has matched their speed and is closing in. Now let's assume the cop drives fairly safely and only gets up to 83mph. With a 10mph speed advantage the cop is closing the distance at about 0.16 miles per minute, so it will take them about 1.25 minutes to make up the entire 0.2 mile gap. At 73mph the speeder will cover another 1.52 miles in 1.25 minutes, so add that distance to the original 0.2 mile advantage and the cop will catch the speeder about 1.72 miles down the road. Obviously this isn't perfect, the cop might be a bit slower to start the chase or might be willing to drive faster to catch a speeder and fill their ticket quota, the speeder might slow down when they see the cop approaching or when they go past the car parked on the side of the road and not travel the whole distance at 73mph, the distance probably isn't exactly two miles because it's measured from exit number to exit number, etc. But overall the ~2 mile claim is fairly reasonable.

TL;DR: your friend got a well-deserved speeding ticket and the cop's report is probably accurate.

 Goremaul wrote:
Well, 144mph is what they'd reach in order to travel the 2km in the same time that the other car does. In reality, he might reach 120 or so, and the other car might start to slow down to stop. That makes the math a bit more complicated though.

That's also assuming a constant rate of acceleration. Since engine don't really do that, you'll have periods of time with faster and slower accelerations, which will mess with the top velocity.

 DarkLink wrote:

 Goremaul wrote:
Well, 144mph is what they'd reach in order to travel the 2km in the same time that the other car does. In reality, he might reach 120 or so, and the other car might start to slow down to stop. That makes the math a bit more complicated though.

That's also assuming a constant rate of acceleration. Since engine don't really do that, you'll have periods of time with faster and slower accelerations, which will mess with the top velocity.

Yeah we realized the deceleration would likely mess with the calculation a bit. I think it was more of a "I can't immediately back of the napkin this formula" intrigue than anything else for us.

It's a trick question. He'll never catch up to you because he just keeps halving the distance.

 AgeOfEgos wrote:

 Goremaul wrote:
I think the cop would have to have an acceleration of ~88.8 miles/hour.

The car travels the 2 miles in ~1.643836 minutes (2 / 73).

Cop car's acceleration:
d = vt + (1/2)at^2
d = 2 miles
v = initial velocity = 0
t = 1.643836 minutes

If his acceleration was 88 mph, that would mean his top speed would be 144 mph--which doesn't seem right. This is using the acceleration formula;

A = (V2-V1)/Time

88 mph = (V2-0) / 1.64 minutes
88 * 1.64 = V2
V2 = 144

Or am I missing the conversion of mph to miles per minute?

Did you do this wrong? I'm fairly certain you did. You see miles per hour is speed so it's velocity not acceleration. Acceleration is distance divided by time squared. Basically you tried plugging velocity into acceleration. It should be A=(V2-V1)/Time. So A=(88-0)/1.64 minutes or 88 (miles per hour)/1.64 (minutes)=acceleration. Then you do the rest of the math whatever 88/1.64 is (i'm too lazy so i can't bother right now). I think you also might want it to be the same so try going for miles/hour squared so multiply the 1.64 by 60 i think.

The way you did your math you had speed (88 mph) equaling out to speed divided by time which is acceleration. This is wrong. Both sides must be equal in measurements. If one side equals the other they must both be speed, acceleration, time, etc. If you check what you did you'll see your end result is velocity (speed basically) equaling out to 144 mi./h x minutes. If you even did the times to cancel out both hours and minutes then you'd be left with distance. Speed doesn't equal distance so this answer was done incorrectly. Anyway we all make mistakes. Happens to everybody. I don't think anybody told you about planck's fudge factor or something. Some story in physics of a guy that had most of the math right but did one thing incorrectly so a guy shared with him the theory based on correcting his one mistake. The lesson to the class was 'always check your work'.

If i remember calculus correctly to get the anti derivative of something (or whatever it was called) you can reverse the number and by multiplying by half in this case and then squaring the answer. This makes sense for the original thing of 1/2(at)^2. Anyway the point of an anti derivative is going from something like acceleration which is distance divided by time squared into velocity which is distance divided by time.

I hope this isn't going too far over your head.

----------

Ok so i figured out that you guys all had it wrong from the beginning and tried thinking speed was acceleration (for some reason). Avoid the first part of what i said then. Let's take this slow.

As for the original question distance = velocity x time in the shortest version possible. The real thing you have to figure out here is how long it took the other person to travel 2 miles assuming both cars started at the same time. This means you also have to figure out how much time somebody traveling 73 miles per hour goes in 2 miles. 73 miles/hr so 73 (miles=distance)/hour (hour=time) over 2 (miles=distance). Basically this is 2 miles=73 mi./hr x time. Divide speed from both sides. 2 miles/73 mi./hr (multiply by 60 for minutes so 73/60 and then 2/(73/60) )=Time. Multiply miles by hrs./mi. as well as the bottom bit to even it out. Basically the miles crossing each other out leaving hours to minutes so time equals time. It's almost nearly 2 divided by one so it was also nearly two minutes.

Now that we have this we can do the math gibberish that goremaul tried to do. Plug that number into time. Now as we know the original velocity was 0 so the whole (vt) bit of his equation isn't needed. At this point it's just d (2 miles)=(1/2)at^2. acceleration is (distance divided by time squared) and multiplying by time squared eliminates time so we just have distance. All you have to do is check the time i gave you from the first bit and plug that gak in. Then we can figure out the acceleration and we should be set.

My work here should be done. Now i shall take a bow and go to sleep.

By my Maths, assuming normal physics conditions:

Acceleration (a) = 1.48018074 mi/h/s (0.6617 m/s/s)
Final Speed (v) = 145.999911 mi/h (65.2678 m/s)

Again, though, that's in physics world with constant acceleration. As Peregrine said, in the real world it wouldn't be constant the entire time.

After guessing top speed values (hey... I'm bored) I got the following example:

Assuming it takes 10 seconds for the cop to begin pursuit:

Top Speed = 82 mi/h
Time Spent Accelerating = 10.933 s
Time taken to close gap between cars = 76.8715 s (almost 1 second under the maximum amount of time the cop car could take for the example top speed to be correct, so pretty close)
Acceleration = 7.5 mi/h/s