Hi
I need help there is a formula for finding the speed of a car by the skid marks it leaves
What is the eqation and is it straight forward

It's going to depend on a number of factors such as the weight of the car, the type and condition of the tires, condition of the road, etc. But from physics it's pretty easy to show that once the car locks up its wheels in a panic stop the length of the skid marks is proportional to the square of the initial velocity of the car. For example, if it takes 80 feet to stop at 40 MPH, then it takes 320 ft to stop from 80 MPH. These distances do not include reaction time - just the stopping distamnce after brakes are applied

According to Oregon Trucking Associations: Stopping Distances For Cars vs. Trucks (http://www.ortrucking.org/stopping.htm) the typical stopping distances for cars is about 80 feet at 40 MPH, and 165 feet at 55 MPH. For trucks its longer.

Here's a chart comparing the stopping distance of various makes and models:
Physics - Car Braking Distances (http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Kinematics/BrakingDistData.html)

The only thing to add is that I think these numbers are for skid marks that stop short of impact. If the situation is that there are skid marks that end in a crash, then this process won't get an accurate answer. You'll have to somehow compensate for the length of skid marks the car DIDN'T make due to the impact.

ebaines,
I think I missed something in your formula. 40^2 is 1600; 80^2 is 6400. But, 40^2/20 is 80 and 80^2/20 is 320?

When working with accident investigation, you also take into consideration the surface of the road, was it wet, was it ice.
There can be a variable for the age and condition of the tires.
You have the actual skid marks, but they will have to lead to the Initial point of impact. And that damage is also reviewed.

But for a general guidelines there are feet per mph

ebaines,
I think I missed something in your formula. 40^2 is 1600; 80^2 is 6400. But, 40^2/20 is 80 and 80^2/20 is 320?

Yes. 80 MPH is two times 40 MPH, so the distance required to stop will be 2^2 = 4 times longer (distance is proportional to the square of the initial velocity). So if it requires 80 ft to come to a stop from 40 MPH, then it takes 320 ft to come to a stop from 80 MPH.