Deceleration force Vs. distance question

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Hi Everyone.

I have a question for you all that is most definitely not homework - Its over 20 years since I was at school and sadly I can't remember very much of what our maths teacher told us... hence the reason I'm here!

I work for a company that does off-road vehicle modifications, these modifications add to the weight meaning that the braking distance is increased.

For many years we have been testing the vehicles using a brake tester that reports the deceleration force in "Negative g"

However we now have a customer who wants us to report the stopping distance - Easy if you can measure it which is what we've been doing.

However, I can't help thinking there must be a relationship between the vehicle's weight, starting speed, the "negative g" deceleration and the stopping distance.

So, if the vehicle weighing 3900kg is traveling at 40km/h and we achieve a deceleration force of 0.9g, what will the stopping distance be? How can I calculate it for other vehicle weights?

Answers in metric please as I'm in Australia! I'm not an engineer and I don't have many qualifications so please excuse the lack of an attempted solution.
 
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As force = mass * acceleration, by putting values in this equation you can easily find the deceleration(F/M).
Now you have initial velocity(u),deceleration(a) and the final velocity(v) of the vehicle, zero(as the vehicle stops ).
Use the equations of motion to find the answer.
I will suggest you to use v^2-u^2=2as.(here 's' represents distance).Put the numeric values in this question and solve this equation for variable 's'.
Use SI units in calculation (force should be in Newtons during calculation)
 
You don't need the force. You already have the acceleration.
To find acceleration in m/s^2 multiply the "g-value" by 9.8 m/s^2.
Then use the formula given above to find stopping distance.
d=v^2/(2a)

where d will be the stopping distance in meters, v the initial speed in m/s and a the acceleartion in m/s^2.

To find speed in m/s divide the value in km/h by 3.6.
For example, 40 km/h is about 11 m/s.

This will work OK only if the acceleration is constant for the duration of the motion.
You also assume that the braking force (or deceleration) measured by your lab method is the same as the one experienced by the car on the road. It may not be the case.
So some experimenting may be necessary.