Recent content by Probie

This thread can be marked as solved. Thank you very much again Chi Meson. You people are the greatest in the world. Thanks.

Thank you very much Chi Meson, I was not even close, man did I make a mess. I really appreciate you pointing me in the right direction. Now that I see how it is done, I will not forget this lesson. Thanks again ~ Probie

Homework Statement This is not homework, I am too old for school. I am trying to teach myself accident reconstruction. The equation I have for impact utilizes crush points. The crush points reveals an energy of 479609.6 to crush the car, so what I am trying to figure out is the speed of the...

I am sorry I thought you wanted me to change the weight to mass which is why I divided the weight by gravity. As for the velocity, I got it by using the skid to stop equation. S = sqr (254 µ d )) S = sqr (254 * .7 * 21)) S = sqr 3733.8 S = 61.10 km/h * .2777 = 16.96 m/s I got...

Thank you very much for your responce and your vote of encouragment. No, when I try it now I loose joules .5 * (13000 / 9.81 ) * V² = 191037.98 Joules I do not know why it is that I have now lost 62.02 joules w = FD = µmgD .7 *1325.178389 * 9.81 * 21 = 191100 joules ( now I...

I am trying to prove that the skid to stop formula works with vehicles of different weight, with the use of kinetic energy. I think I have proven this, but if someone would like to check this over to make sure I am correct I would appreciate it. 13000 kg vehicle leaves skid marks 21 meters...

solved myself Thanks, but I proved this one myself, with the use of kinetic energy. But if someone would like to check this over to make sure I am correct I would appreciate it. 13000 kg vehicle leaves skid marks 21 meters long, the drag factor is .7 13000 * 21 * .7 = 191100 joules V =...

Homework Statement Is it possible that a vehicle weighing 2000 kg and a vehicle weighing 13000 kg can both use the same skid to stop formula?? Does the difference in weight not come into account? Homework Equations Where : S = speed in km/h µ = .75 d = skid distance S = 15.9...

Thanks to both of you, the information is very helpful. Here is the complete situation. I was asked to check on this for work. A vehicle is stopped on the highway at a laneway. Another vehicle crests a hill at a speed of 80 km/h. Is there enough time for the second vehicle to stop before...

Okay, let's say you have to give an opinion on the following. A car is approaching the rear of a stationary car. And you want to know if it will crash into the car or if there is enough distance to stop. You need to know the deceleration rate of the car. You do not want it to skid to a stop...

I do not understand vehicle deceleration. The Attempt at a Solution d = µg 6.687 = .7 * 9.81 d = 6.687 m/s My question is...is it always figured out this way when dealing with vehicles.

Thanks again cristo...this is a lesson I will never forget...thank you very much. ~ Probie

Yes, that is exactly what I needed to know. It sure made a difference in my answer...now I come up with E = 1726.66675 I sure do hope this is right. I am not a student and there is no way to check my findings. Cristo...thanks very much for the prompt reply, this is very important to me...thank you.

Can someone please tell me what I am suppose to do with c1c2 and 2b in the following equation. I do not know if I am to mulitply, divide, subtract or add. E= L * (( a /2 ) * ( c1 + c2) + b / 6 * ( c1² + c1c2 + c2² ) + ( a² / 2b)) Where: L = 1 a=375 b=45 c1=.33 c2=.33 My answer comes...

Acceleration is ft/sec² is it the same for meters even though in metric it is written as m/s/s? That is my question. Maybe I am just to lame to get this stuff. But I am confused.