# A car travels around an unbanked highway curve of radius 50m. The coefficient of static friction is 0.6. The maximum speed that the car can travel without skidding is A. 22.1m/s B. 23.4m/s C....

A car travels around an unbanked highway curve of radius 50m. The coefficient of static friction is 0.6. The maximum speed that the car can travel without skidding is

A. 22.1m/s

B. 23.4m/s

C. 26.5m/s

D. 17.1m/s

E. not enough information given

Correct answer is D

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### 1 Answer

If a car is traveling on an unbanked curved highway, the force of friction is directed towards the center of the curve. This is the centripetal force that keeps the car moving along the curve.

According to the second Newton's Law, the net force equals mass times acceleration. The net force is the sum of friction (`vecF `) , gravity, and the normal force:

`vecF + m*vecg + vecN = m*veca ` ,

where `veca ` is the centripetal acceleration.

The friction force and the acceleration lie on the plane of the highway (horizontal), while the gravity and the normal force are directed along the vertical line (gravity - down, and normal force - up.) So the horizontal and vertical components of the above equation are, respectively:

F = ma

-mg + N = 0; from here, N = mg

The friction force is related to the normal force by the equation

`F=mu*N = mu*mg ` .

The magnitude of the centripetal acceleration is related to the speed of the car and the radius of the curved highway:

` a = m*v^2/R`

Substituting expressions for F and a in the first equation results in

`mu*mg = m*v^2/R `

The mass cancels and from here, speed can be found:

`v^2 = mu*gR `

`v = sqrt(mu*gR) = sqrt(0.6*9.8 m/s^2*50 m) =17.15 m/s `

This is approximately equal to choice D.

See the reference link for the more general case of the car on a bank roadway. Note that when roadway is unbanked, angle `theta ` is zero and the friction force becomes equal to the net force, just like in this case.

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