Answer:
12.6332454263 m/s
Explanation:
m = Mass of car
v = Velocity of the car
[tex]\mu[/tex] = Coefficient of static friction = 0.638
g = Acceleration due to gravity = 9.81 m/s²
r = Radius of turn = 25.5 m
When the car is on the verge of sliding we have the force equation
[tex]\dfrac{mv^2}{r}=\mu mg\\\Rightarrow v=\sqrt{\mu gr}\\\Rightarrow v=\sqrt{0.638\times 9.81\times 25.5}\\\Rightarrow v=12.6332454263\ m/s[/tex]
The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s
