A banked circular highway curve is designed for traffic moving at 57 km/h. The radius of the curve is 216 m. Traffic is moving along the highway at 43 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road

The minimum coefficient of friction must be 0.067

Explanation:

In order to allow the cars to take the turn without sliding off the road, the Centripetal Force must be equal to the frictional force between the tires  of the car and road.

Centripetal Force = Frictional Force

but centripetal force = mv²/r

and friction force = μR

mv²/r = μR

where,

m = mass of car

v = speed of car = 43 km/h = 11.94 m/s

r = radius of circle = 216 m

μ = coefficient of friction = ?

R = Normal Reaction = Weight of Car = mg

Therefore,

mv²/r = μmg

μ = v²/gr

μ = (11.94 m/s)²/(9.8 m/s²)(216 m)

μ = 0.067

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explanation:

i think the answer is a, hope this is right, that would be my best guess.