How long does it take a truck to stop & is it criminal if he doesn't?

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On Wednesday, July 16, 2014 1:48:38 PM UTC-4, Ed Pawlowski wrote:

That's for sure. I wonder if anything else was going on, like the passenger giving BS to the cop. It still would not justify giving him a DUI.

I have problems with cops arresting people for DUI under those circumstances too. If I were the cop, and I found someone safely pulled over, sleeping in the car, in a reasonable place, I'd give them a pass. If they fell asleep at a light, or were parked somewhere they clearly did not belong, that would be different. I think it's better to send a message that if you're drunk and make a mistake, but realize it, you have a safe way out. If you know you're going to get a DUI for going to a rest stop, it really leaves the best option for everyone off the table.
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wrote:

Not DUI. but care and control while under the influence. If you are going to "sleep it off" in your car, make sure you hide the key OUTSIDE of the vehicle. A friend's wife was sitting in the car after an accident and was charged with "care and control". If she had given someone else thekeys, or stayed out of the car, they would not have been able to charge her and make it stick. No proof she was the driver, and no proof she had been drinking before the accident.
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On Wednesday, July 16, 2014 5:44:59 PM UTC-4, snipped-for-privacy@snyder.on.ca wrote:

One has to wonder how you can be charged with "care and control while under the influence" with no proof that you had been drinking before the accident. Did she have a bottle with her and claim that she had been drinking *after* the accident?
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wrote in message

Elementary Watson!
SH
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I haven't read all 98 posts, but this doesn't seem right. If G increases, that is, a steeper hill, d increases, and that is fine.
But if f increases, the formula shows that d also increases, and that seems clearly wrong. The higher the coefficient of friction, that is, the more friction there is between the road and the tires**, the sooner the truck stops. A very low f would be like melting ice. A very high f would be like a road made of big rocks, maybe.
At any rate f and G are parellel in the formula, even though high values of each have the opposite effect. Maybe there should be a negative sign somewhere.
**And what about between the brake pads and the rotors? Or are we assuming the wheels are locked up? I didn't think that was the case.

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On Saturday, September 6, 2014 6:12:24 PM UTC-4, micky wrote:

You also must have been MIA, this is an old thread.
If G

Actually if either G or f increase, then d will *decrease*. Both G and f are in the denominator.
The higher the coefficient of friction, that is,

I think there is a negative sign, depending on whether the truck is going up the grade or down it. For the formula to be correct, it would appear that a positive grade is one where the truck is going uphill. If the grade is negative, ie truck going downhill, then when the grade cancels out the friction, the stopping distance becomes infinity. Beyond that, the formula doesn't work. If the downhill grade exceeds the friction, it would give a negative stopping distance, but that isn't right, the truck would never stop, so it doesn't work for neg denominators.

I would think the formula is based on the wheels being locked up. How close it is to reality, even then, IDK.
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wrote:

You are correct. I pointed that out in one of the other posts in this thread. I don't blame you for not reading them all.

The use of the formula relies on the user understanding that the correct sign must be used for GRADE.

The formula only deals with the friction, f, at the tire/road interface. The formula can be used to solve for any of it's variables based on knowing the other variables and plugging them in to get a solution. It also needs to be kept in mind that the way it's often used, and how its being used here, one simply makes an assumption that is said to be the f for the road. That's a bit simplistic because there is no "f' for the road. f only has meaning relative to what ever material is the other half of the interface. In this case it's a tire. And different tires can produce different f's, sometimes significantly different. Also, as you mention, there is the issue of whether the brakes are locked. If they are locked and the tires skidding you have a different f value then what you would get/use if the tires were not skidding. F also varies with the speed of the things moving against each other at the interface, so the f value is an average of all the f's as the speed goes from the starting speed down to zero.
Or are we

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