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, [tex]\mu = 0.200[/tex].
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
[tex]\dfrac{m \times v^{2}}{r} = \mu \times m \times g\\\\\\\dfrac{v^{2}}{r} = \mu \times g\\\\v =\sqrt{\mu \times r \times g}[/tex]
Solving as,
[tex]v =\sqrt{0.200 \times 75.0 \times 9.8}\\\\v =12.1 \;\rm m/s[/tex]
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:
https://brainly.com/question/14249440
