# Simple word problem that I can't figure out.

1. Nov 25, 2004

### dmlindsay90

I hope that I construct this problem correctly. I am trying to....um, help a friend fight a speeding ticket. Thanks in advance for any help.

The posted speed limit was 25 MPH. The officer was stopped at an intersection as I passed by him on the perpendicular and "estimated" that Iwas traveling "in excess of 35 MPH." He stated that by the time HE reached the next intersection, which is ~300 feet from his starting point, he had "paced" me at ~35 MPH and "activated his emergency lights and signalled" me to stop. I have already determined that if I was traveling at 37 MPH it would equal 54.25 ft/sec and therefore the distance would have been covered in about 5.42 seconds. I have also assumed that he would have had to pace me for SOME period of time (2-3 seconds) leaving only 2.42 to 3.42 seconds in which to catch up to me from the time that I crossed his path. What I am having trouble determining is:

1. How rapidly would he have had to accelerate in order to catch up to me ?
2. To what velocity would he have had to accelerate in order to catch up to me?

2. Nov 25, 2004

### Tide

Do I understand you correctly that you intend to use the fact you were speeding as a defense against a charge of speeding? Obviously, the faster you were going the faster he would have had to accelerate. However, he can pace you without having overtaken you. That can easily be done over a wide range of vehicle separations. Also, be aware that your calculations won't necessarily receive a warm welcome unless you can qualify yourself as an "expert witness" which is highly unlikely.

3. Nov 25, 2004

### dmlindsay90

Sort of. What I am trying to show is that there would not have been enough time for him to catch up to me AND pace me for a couple of seconds by the time he decided to pull me....um, I mean my friend, over.. Since I made my first post I have done some more research and found a report from the Michigan State Patrol that tested numerous vehicles and the fastest one accelerated from 0 to 30 in 3 seconds and from 0 to 40 in 4.42 seconds. Since it would have only taken about 5.5 seconds to travel the 300 feet, I would like to prove that I COULDN'T have been going "in excess" of 35 MPH.

If he began following me .25 seconds behind me, by the time he reached 35 MPH I would have traveled almost 190 feet and he wouldn't have made up any ground yet. That leaves only about 100 feet for him to:
a. Close to gap enough to me to pace me
b. Equalize our speeds and look at his speedometer, and
c. Light his Christmas tree.

It seems obvious to me that his "estimation" was not very accurate, but I haven't been able to come up with the actual method to prove my hypothesis.

4. Nov 25, 2004

### DaveC426913

I have had a car pace me from what must have been several hundred yards distance. I don't know how they can tell at that distance over so short a time.

It seems to me that it's pretty much exclusively their word on it. There's no record of any distacnes or speeds. They could say whatever they want.

5. Nov 25, 2004

### dmlindsay90

Right, but if what they SAY can be disproved by scientific fact, then their word becomes worth less (or even worthless).

6. Nov 25, 2004

### KingNothing

Alright...first of all, some of your assumptions are incorrect. For example, he wouldn't need to 'pace' you for 2-3 seconds. In fact, all he would have to do is accelerate up to 32 mph (cops pull people over @ 7mph over) and see that he isn't gaining ground, then guess at your speed. So to see if he could determine if you were speeding, the question becomes "Could he accelerate to 32 mph in 300 feet".

So take the best acceleration from your statistics, 10 mph/s, and figure it out. Figure out the distance it takes him to get from 0 to 32. Now, based on the acceleration, it would take him 3.2 seconds.

Distance = 1/2 * accel * time^2
Distance = 1/2 * 10 * 3.2^2

Last edited: Nov 25, 2004
7. Nov 26, 2004

### KingNothing

1.75
61.25

Where do you get 190 feet? If he accelerates at 10 mph/s, it would take him 3.5 seconds plus the .25= 3.75. In 3.75 seconds you would have traveled 3.75*35=about 202 feet.

That gives him 98 feet to overtake you.
Since he would have bee accelerating for 3.5 seconds to reach 35 mph, he would have covered a distance of...
d=1/2*a*t^2
about 61 feet, leaving 202-61 or 141 feet between you and him, and 98 feet left until the next intersection.

S now the question is, at his acceleration, how long would it take him to cover that extra 141 feet?

141 = 1/2*10*t^2
141=5t^2
28.2=t^2
t=5.31 seconds

in 5.31 seconds, you would have covered another 288 feet, leaving another -190 feet to go to the next intersection. This essentially proves you correct, but again it's based on assumptions. If he was to pace you right as you are crossing the next intersection, (assuming he paces instantly) he would have to do it from a distance of about 125 feet. I want to see you make that calculation.

8. Nov 26, 2004

### lordinfamous

radar travels faster than the time he would have needed to pace your speed.
Your calculations no matter how good they might be will not stand in traffic court and will be seen as a mockery. but good luck with it anyway! No pun intended.

9. Nov 26, 2004

### dmlindsay90

What is the "1/2" that you have in your formula? And how do you go from 10 mph/s to feet traveled w/o converting it?

Using A = (delta)V / T to find the acceleration of the police car I get:

A = (51.33 fps - 0 fps) / 3.6 sec (time I interpolated it takes to go from 0 - 35)
A= 12.83 fps^2

Then plugging that figure into this formula,
D = A * T^2 I get:
D = 12.83 fps * 12.96 sec
D = 166.28 feet

Now, I don't know it I am doing this correctly, but it seems more logical. To think of a vehicle accelerating from 0 - 35 MPH in only 50 feet seems way too quick.

10. Nov 30, 2004

### ricemike

give it up
pay the ticket
stop playing smart games with smarter people

11. Dec 1, 2004

### dmlindsay90

I assumed this was a forum about physics (I am smart enough to read the website name). So I asked a question that I am having trouble conceptualizing which can be explained using a few simple formulas. If you have any input regarding the question, I would like to hear it, but if all you can do is provide sarcastic (and worthless) legal advice, you can keep it to yourself.