The coefficient of car tires on pavement is 0.91 ± 0.10. This tabdelimited text file contains the stopping distance data for 123 cars tested by Road & Track magazine in 1998. Two initial speeds were used: 26.817 m/s (60 mph) and 35.756 m/s (80 mph). Use the data in this file and your favorite data analysis software to determine the coefficient of friction of car tires on pavement.
Start with Newton’s second law of motion.
∑ F = m a
A stopping car is acted upon by three forces: weight pointing down, normal pointing up (we’ll have to assume the test track is level), and friction. (Of course, there’s also aerodynamic drag, but worrying about that force would be a waste of time.) Weight and normal cancel out since the car is neither accelerating up nor down. Friction is therefore the net force acting on a braking car.
ƒ = ma
Replace ƒ with its classical formula.
µN = ma
Earlier, we assumed (quite sensibly) that the test track would be level, which means that normal equals weight (W = mg).
µmg = ma
Work the magic of algebra and solve for the goal of this problem — the coefficient of friction.
Great, but what is a? Go back to the good old days when you learned the equations of motion. Pick the one that doesn’t involve time and solve it for acceleration.
_{ }v^{2} = 
v_{0}^{2} + 2a∆s 


_{ }^{ }a = 
v^{2} 

2∆s 
Substitute this expression into the previous one.
Take all the numbers in roadtestsummary.txt and run them through this final equation. These results are given in roadtestsummarysolution.txt. (Note: I used g = 9.8 m/s^{2}, but one could also use the value of standard gravity g = 9.80665 m/s^{2}.) Using the mean of these 246 trials as the value and the standard deviation as the uncertainty yields the following answer.
µ = 0.91 ± 0.10
This problem can be found on the practice problems page of the section on friction in The Physics Hypertextbook.