Posted by Noah on .
To test the performance of its tires, a car
travels along a perfectly ﬂat (no banking) circular track of radius 120 m. The car increases
its speed at uniform rate of
at ≡
d v
dt
= 2.99 m/s
2
until the tires start to skid.
If the tires start to skid when the car reaches
a speed of 18.2 m/s, what is the coeﬃcient of
static friction between the tires and the road?
The acceleration of gravity is 9.8 m/s
2

AP Physics 
Damon,
a = 2.99 m/s^2 is irrelevant I suspect
Ac = v^2/r = 18.2^2/120 = 2.76 m/s^2
m * Ac = friction force = m g mu
2.76 = 9.8 mu
mu = 2.76/9.8 = .282 
AP Physics 
Anonymous,
Ac = v^2/r = 18.2^2/120 = 2.76 m/s^2
at ≡d v/dt= 2.99 m/s2
sqrt(2.76^2+2.99^2)=4.069 m/s^2
μs=(amax)/(g)=4.069/9.8=.4152