The radius of curvature of a highway exit is r = 69.5 m. The surface of the exit

Question

The radius of curvature of a highway exit is r = 69.5 m. The surface of the exit road is horizontalIf the static friction between the tires and the surface of the road is s = 0.623, then what is the maximum speed at which the car can exit the highway safely without sliding?, not banked. (See figure.)

If the static friction between the tires and the surface of the road is s = 0.623, then what is the maximum speed at which the car can exit the highway safely without sliding?

Explanation / Answer

Ans:-

Given data :r = 69.5m s = 0.623

Equate all force

Centrifugal force = friction force   

Fc = s *Fn

smg = mv^2 / r
v = sqrt(ugr) = sqrt(0.623*9.80*69.5) = 424.3 m/s

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