quote:

Originally posted by Flyboy217 60 kph = 16.66 m/s 45 ft = 13.7 m v^2 = 2ax a = 10.1 m/s^2 That's approximately an acceleration of -1g. It's very reasonable to believe, as quick cars can achieve a positive acceleration of 1g.

oo, yeah, listen to him. hes smart

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quote:

Originally posted by torontotrance so 10.1m/s = how many km per hour?

Not 10.1 m/s. 10.1 m/s^2, an acceleration. This is a reasonable rate for a car to decelerate. Nothing else (momentum, mass, etc.) needs to be considered.

*edit* Consider it this way. From a standstill, if you were to accelerate at 10.1 m/s^2, it would take you 2.65 seconds to get to 60 mph. That's very quick acceleration, but not neck-snapping. If he was hard on his brakes, it's feasible that he could do the reverse (40 mph to 0 using that deceleration). Double-check me though. Cross reference deceleration rates on the internet.

Last edited by Flyboy217 on Jul-02-2004 at 04:39

quote:

Originally posted by Flyboy217 Not 10.1 m/s. 10.1 m/s^2, an acceleration. This is a reasonable rate for a car to decelerate. Nothing else (momentum, mass, etc.) needs to be considered. *edit* Consider it this way. From a standstill, if you were to accelerate at 10.1 m/s^2, it would take you 2.65 seconds to get to 60 mph. That's very quick acceleration, but not neck-snapping. If he was hard on his brakes, it's feasible that he could do the reverse (40 mph to 0 using that deceleration). Double-check me though. Cross reference deceleration rates on the internet.

http://www.skytran.net/09Safety/10sfty.htm

As you can see, up to 60 g's is considered "safe." We're dealing with ONE g. From 30 mph to 0, they list 1 foot as a safe stopping distance (sounds absurd to me, but hey). Mind you, this says nothing about the car (it would probably become junk at this rate), but a human wouldn't feel much at 1g.

quote:

Originally posted by torontotrance so if the skid marks say it took him 45ft to stop and he says he was going 60kph, is there anyway to figure it out via some formula if the dude is telling the truth.

Ok, d = v*t + (1/2)*a*t^2 is your basic physics formula we're gonna use here.

The only problem is, we don't know how long it took for him to lay the 45ft of skid marks. So we'll swap crap around to get rid of the time element in the second part of that equation.

a = v/t so t = v/a

swap that into the second bit and we get:

d = v*t + (1/2)*a*(v/a)^2

stuff cancels and we are left with:

d = v*t + (v^2)/(2a)

so, in this equation:
d = braking distance
v = the speed the guy was going before he started braking
t = the idiot's reaction time
a = rate of decelleration

if we give this guy the benefit of the doubt, and say he has some kickass reaction time, t = 0.3 seconds
he says he was going 60km/h --> 17m/s approximately
and lets say his tires are ok and he has decent brakes, so he's decellerating at 1g = 9.81m/s^2
(It is unlikely that he was slowing down at this rate, since his tires were sliding on the asphalt).

So, plug that stuff in and you get a theoretical braking distance (from 60km/h to 0km/h), if he was a kickass driver and didn't lock up the brakes of: 20 meters. --> a little over 60 feet.

if you eliminate reaction time from the whole thing, and just look at from the point where his tires locked up to where he hit the car, then the distance is 45ft.

So, since the car didn't end up at 0km/h, millimeters away from the other car, after a 45ft brake lockup, you can be pretty confident that he was going quicker than 60km/h when he locked up the tires.

Sorry for the long post...time to watch Ali G.

quote:

Originally posted by Flyboy217 http://www.skytran.net/09Safety/10sfty.htm As you can see, up to 60 g's is considered "safe." We're dealing with ONE g. From 30 mph to 0, they list 1 foot as a safe stopping distance (sounds absurd to me, but hey). Mind you, this says nothing about the car (it would probably become junk at this rate), but a human wouldn't feel much at 1g.

Better yet, consider this:

http://www.skytran.net/09Safety/02sfty.htm

1g is apparently a good limit for a clean dry road with very good tires. If you factor in reaction time (as N][ick did), then it does start to seem less likely. It's close, but now I'll go with "no" for the guy.

if we knew the type of car he had a quick check to www.kbb.com could tell us the weight
and therefore we can calculate the coefficient of friction on the ground (becuase he slid)

compare it to the normal friction coefficient of the ground and see if its way off

also you could probably tell how fast he was going when he started skidding

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quote:

Originally posted by Photo_bot_2k1 if we knew the type of car he had a quick check to www.kbb.com could tell us the weight and therefore we can calculate the coefficient of friction on the ground (becuase he slid) compare it to the normal friction coefficient of the ground and see if its way off also you could probably tell how fast he was going when he started skidding

The coefficient of kinetic friction between rubber and asphalt is a constant. The deceleration figures are sufficient to describe the braking of a car.

this is all gibberish to me.

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quote:

Originally posted by Flyboy217 The coefficient of kinetic friction between rubber and asphalt is a constant. The deceleration figures are sufficient to describe the braking of a car.

yeah of course thier constant thats why you compare the value you get from your calculations to the book values for the asphault vs rubber

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doesnt the insurance company have a way to figure it out? i know when i got into an accident the cop figured out that the kid was speeding and issued him a ticket on the spot. he started crying and i started laughing (it was one of those evil villian laughs like in cartoons). in the end it doesnt matter because the insurance companies get together and find the best way to screw over their clients.

hope your mom is alright.

http://www.e-z.net/~ts/speedch.htm

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