Answer: 2.74
Explanation:
We can solve this problem using the stopping distance formula:
[tex]d=\frac{(V_{o})^{2}}{2 \mu g}[/tex]
Where:
[tex]d=132.1 m[/tex] is the distance traveled by the car before it stops
[tex]V_{o}=30.7 m/s[/tex] is the car's initial velocity
[tex]\mu[/tex] is the coefficient of friction between the road and the tires
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
Isolating [tex]\mu[/tex]:
[tex]\mu=\frac{2dg}{(V_{o})^{2}}[/tex]
Solving:
[tex]\mu=\frac{2(132.1 m)(9.8 m/s^{2})}{(30.7 m/s)^{2}}[/tex]
[tex]\mu=2.74[/tex] This is the coefficient of friction
