Item 10 A banked curve of radius R in a new highway is designed so that a car traveling at speed vo can negotiate the turn safely on glare ice (zero friction). If a car travels too slowly.then it will slip toward the center of the circle. If it travels too fast, it will slip away from the center of the circle. If the coefficient of static friction increases, it becomes possible for a car to stay on the road while traveling at a speed within a range from Umin to Umax Part A Derive formula for v_(min) as a function of mu _(s),v_(0) and R Express your answer in terms of the variables mu _(5),v_(0),R and appropriate constants. sqrt [0](square ) A2p ? v_(min)=sqrt ((v_(0)^2+mu _(s)gR))

To derive the formula for , we need to consider the forces acting on the car as it negotiates the banked curve.The car is traveling at a speed and is subject to the following forces:1. Gravitational force: , where is the mass of the car and is the acceleration due to gravity.2. Normal force: , exerted by the road on the car.3. Frictional force: , where is the coefficient of static friction.The car is moving in a circular path with radius , so the centripetal force required to keep the car moving in this path is given by: The centripetal force is provided by the combination of the gravitational force and the normal force: where is the angle of the banked curve.The normal force can be expressed as: Substituting this into the equation for the centripetal force, we get: Simplifying, we have: To find the minimum speed , we need to consider the case where the car is traveling at the minimum speed that allows it to stay on the road. In this case, the frictional force is at its maximum value, which is . The car is just able to maintain its circular path, so the centripetal force is equal to the sum of the gravitational force and the maximum frictional force: Simplifying further, we get: Taking the square root of both sides, we obtain: Therefore, the formula for as a function of , , and is: