What is the maximum speed at which a car can safely travel around a circular track of radius 75.0 m if the coefficient of friction between the tire and road is 0.200?
a) 3.87 m/s
b) 12.1 m/s
c) 15.0 m/s
d) 147 m/s

The car can turn through the circular track with a maximum speed of 12.1 m/s. Hence, option (B) is correct.

Given data:

The radius of circular track is, r = 75.0 m.

The coefficient of friction between the tire and road is, .

When the car moves around a circular track, then for safe turn through the track it is necessary to have the value of frictional force and centripetal force in a balanced amount. Therefore,

Fc = Ff

Solving as,

Thus, we can conclude that the car can turn through the circular track with a maximum speed of 12.1 m/s. Hence, option (B) is correct.

Learn more about the centripetal force here:

link

B) 12.1 m/s

Explanation:

Sum of the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of the forces in the radial direction:

∑F = ma

F = m v² / r

Nμ = m v² / r

Substituting and solving for v:

mgμ = m v² / r

gμ = v² / r

v = √(gμr)

Given that μ = 0.200 and r = 75.0 m:

v = √(9.81 m/s² × 0.200 × 75.0 m)

v = 12.1 m/s

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray