Question

A concrete highway curve of radius 70 m and is banked at an angle of 30o. imagine a car driving around the turn at a constant height. on a rainy day, the concrete will be wet and the coefficient of static friction is s = 0.3. we want to find the maximum speed at which a 1100 kg rubber-tired car can go around this curve.

Answer 1

The maximum speed at which a 1100 kg rubber-tired car can go around a concrete highway curve of radius 70 m and banked at an angle of 30° is approximately 152.06 m/s.

Step-by-step explanation:

To find the maximum speed at which a 1100 kg rubber-tired car can go around a concrete highway curve of radius 70 m and banked at an angle of 30°, we need to consider the forces acting on the car. The two main forces are the gravitational force and the frictional force. The maximum speed occurs when the static friction force provides the centripetal force required to keep the car moving in a circle.

The centripetal force is given by the equation Fc = mV^2/R, where m is the mass of the car, V is the velocity, and R is the radius of the curve. The maximum static friction force is given by the equation Fsf = usFn, where us is the coefficient of static friction and Fn is the normal force.

By setting Fc = Fsf, we can solve for V to find the maximum speed. Plugging in the values, we get:

V^2/70 = 0.3 * 1100 * 9.8

V^2 = 23100

V = √23100

V ≈ 152.06 m/s