Answer:
vmax=72.59m/s
Explanation:
To solve this problem we use
[tex]v_{max}=\sqrt{Rg\frac{sin\alpha +f_{f}cos\alpha }{cos\alpha-f_{f}sin\alpha}}[/tex]
where R is the radius of the curve, g is the gravity constant, α is the degree of superelevation, ff is the friction coefficient. By replacing we have
[tex]v_{max}=\sqrt{(800+6)(32.2)\frac{sin(0.06)+0.14cos(0.06)}{cos(0.06)-0.14sin(0.06)}}=72.59\frac{m}{s}[/tex]
where we have taken into account that R=800+6 due to the 12-feet of the Lanes, but we take one half, because we assume the car is at the center of the Lane.
I hope this is useful for you
regards
