| Author |
Message |
Mark Iverson
| | Posted on Monday, 15 December, 2003 - 05:54 pm: | |
Hello,
I received a speeding ticket the other day and to build my defense I need the following information and do not possess the mathematical ability to figure this out. Could anyone please help me? I was cited for going 41 MPH in a 25 MPH zone. The officer said that he picked me up on his radar going 41 MPH almost immediately after I came to a stop and preceded. I probably advanced less than 50 feet before he clocked me at 41 MPH. I want to show the judge that it would be impossible to reach a speed of 41 MPH in such a short distance. I know this about my car which is a 1992 Honda Accord. It goes 0 - 60 MPH in 9.8 seconds. It does the 1/4 mile in 17.4 seconds. So with that we know that the car can averages 75.86 FT per SEC at 9.8 SEC to go 0-60 MPH and we also know that the car traveled 743.45 FT to reach 0 - 60 MPH in 9.8 seconds. What I want to know is how many feet would the car have to travel to reach a speed of 41 MPH and if possible how many seconds would it take. I know we are probably missing some acceleration info here, but anything logical and close that I could show or prove to the judge with math would be a great help for my defense. I am not guilty and the officer refused to show me his radar reading to prove to me I did this. Thanks |
Chris Tynan
| | Posted on Monday, 15 December, 2003 - 06:34 pm: | |
60 Miles is approximately 100000 metres. So 60 mph = 100000 metres per hour = 28 metres per second roughly. So the acceleration is approximately 2.8 ms-2. Our final speed is 41 miles per hour = 18.8 metres per second. So using v2 = u2 + 2as we have (18.8)2 = 2.(2.8).s. So s is at least 63 metres. |
Shahnawaz Abdullah
| | Posted on Monday, 15 December, 2003 - 06:50 pm: | |
I agree with you - it is quite impossible you were going at that speed.
This is how I would personally argue it-
We assume constant acceleration
You went from 0 - 41 MPH which is 0 - 18.8 m/s
This was in 50ft = 15m
but v^2 = u^2 + 2as which implies that the acceleration is
(v^2)/2s = (18.8*18.8)/(2*15) = 11.78 m/(s^2)
HOWEVERRRR
we know the car can do 0 - 60 MPH (0-28m/s) in a min time of 9.8s
using v = u + at, we can say that 28/9.8 is its max acceleration is 2.9 m/(s^2)
but for you to have gone from 0 - 41 in 50 ft, you needed an acceleration of 11.78 m/(s^2) which would be quite impossible for your car.
HOWEVER if you hadn't travelled 50ft, then all this falls apart :P |
James
| | Posted on Monday, 15 December, 2003 - 06:52 pm: | |
Also/alternatively, using your information, 41mph = 18.8m/s
50 feet is about 15 meters, so
v2 = u2 + 2as
18.82 = 0 + 30a
a = 12 ms-2
compared to 2.8 ms-2
So to reach 41 mph after 50 feet, your engine would have to be working about 4 times harder than it is capable of. |
James
| | Posted on Monday, 15 December, 2003 - 06:54 pm: | |
Heh, just written pretty much the same as you Shahnawaz...never mind. |
Graeme McRae
| | Posted on Monday, 15 December, 2003 - 07:18 pm: | |
I guess a number of us find this topic interesting! Here is my reply, pretty much along the lines of Shahnawaz and James...
60 MPH = 88 feet per second.
So the acceleration in feet per second is 88/9.8 = 8.98 ft/s2.
In most cars, the acceleration is not constant over the range 0-60 MPH, but if it were then you could solve the equation
50 = (8.98/2)t2
to find the time, t, it takes the car to reach the 50-foot mark. That time is about 3.3 seconds. In 3.3 seconds, your speed reaches 8.98*3.3 = 30 feet per second, or about 20.4 MPH.
While you have the accelerator pedal fully depressed, a Honda Accord stays in first gear until about 40 MPH, and so the acceleration is much higher than the average over 60 MPH. But the acceleration would have to be greater than that of gravity for you to reach 41 MPH in just 50 feet. If the cop was right, and if you really stopped at that stop sign (you did stop, right?) then you would have gone from zero to 41 in just 1.66 seconds -- quite an exhilarating ride!
If you can prove that your speed was clocked just 50 feet from the stop sign, and you can prove that you stopped at that stop sign, then you can prove that the acceleration your car experienced during that 50 feet far exceeds the capacity of a standard Honda Accord. Here are the equations:
Let a be the acceleration, in feet per second, of your car.
Let t be the time, in seconds, that elapsed from the stop sign to the 50-foot mark, where you were going 41 MPH, or 60.1333 ft/s.
These two equations relates a and t, given that you accelerated smoothly from 0 to 60.1333 ft/s, and traveled 50 feet doing so:
(1) 50 = (a/2)t2
(2) a = 60.1333/t
Since you were in first gear the whole way, and the torque generated by the Honda Accord is fairly constant over that range, the smooth acceleration is a reasonable approximation.
Substituting 60.1333/t in place of a in the first equation, we get
50 = 30.0667t
so t = 1.66 seconds
From the second equation, we find a = 60.1333/1.66 = 36.16 ft/s2, which is quite a bit greater than the acceleration of a Honda Accord.
Now, as a practical matter, you may find it difficult to prove that the reading was taken just 50 feet from the stop sign -- that's probably no more than the width of the street you crossed. If the reading were taken, say, 50 yards instead of 50 feet from the stop sign, then your whole case is toast. I think you stand a much better chance of acquittal if you argue that the cop didn't take a measurement of your car at all. Your evidence is that he did not show you the reading. Good luck! |
Mark Iverson
| | Posted on Monday, 15 December, 2003 - 07:25 pm: | |
Shahna or anybody,
Can you show me how to convert 0 - 18.8 m/s to 0 - 41 MPH? Also convert everything to FT/SEC. The judge won't understand meters and I would like to explain these calculations to his way of thinking. What are the v^2 = u^2 + 2as symbols and v = u + at. Also, I don't know how far I traveled exactly before he clocked me, I just estimated 50 FT it could have been 20 or even 100. Thanks. |
Graeme McRae
| | Posted on Monday, 15 December, 2003 - 07:35 pm: | |
To convert from MPH to ft/sec, multiply by 88/60.
To convert from ft/sec to MPH, multiply by 60/88.
To convert from ft to m (or ft/sec to m/sec), multiply by 0.3048.
To convert from m to ft, divide by 0.3048.
If you're not sure that the reading was taken 50 feet from the stop sign, then you need to put an upper bound on how far it could have been, and -- this will be the hard part -- get the parties to stipulate this upper limit.
If the upper limit is, say, 150 feet, then you might as well drop this line of defense, because a Honda Accord with good tires on a dry surface can probably accelerate to 41 MPH in that distance. Remember, you're in first gear the whole way, pushing 7000 RPMs at the end of your run. Was the road level? Even a slight downhill grade can shorten the distance it would take your car to reach 41 MPH.
Now, if you'll permit me a personal observation, getting a speeding ticket presents wonderful educational opportunities. The first opportunity is this exploration of physics. If you live in a state (like California, where I live) that lets you go to "traffic school" in exchange for dismissing a speeding ticket, then this will be your second educational opportunity! |
James
| | Posted on Monday, 15 December, 2003 - 08:23 pm: | |
Also, this is the only one i know
1 mile = 1.596 km, so 41 mph = 41 * 1.6 km/h
v2 = u2 + 2as
v = u + at
These are standard laws of motion,
v = final velocity
u = start velocity
a = acceleration
t = time in seconds
s = distance
In this case u is always 0 since you started from rest.
Graeme, in the USA, do you do all your physics using imperial units, or do you convert to metric then back?? I can imagine it gets rather messy. |
Graeme McRae
| | Posted on Monday, 15 December, 2003 - 08:35 pm: | |
In the USA, or anywhere else, I suppose, real physics is done using metric units. But "play physics" involving just two dimensions (time and distance, in this case) can be done in any units without getting too badly tripped up. |
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