Let's talk about a deal of banking. When a car makes a horizontal turn on a curve of radius R while traveling at a velocity V, curve must be bank at an angle theta so that there is no tendency for the car to slide up and down the road. This angle is called a deal of banking. Given the formula tan theta is equals to V squared over RG. Let's proceed to problem number 1. Hacker road has a radius of 25 meter and bank at 20 degree. What is the maximum speed? Given the values of radius is equal to 25 meter at the angle of 20 degree, and the constant acceleration due to gravity is equal to 9.8 meter per second squared. The required on this problem is the maximum speed of V. Given the formula of tan j and theta equals V squared over RG, let's equate the formula to V squared equals Rg tan j and theta to V, is equals the square root of Rg tan j and theta. Then substitute the given value V is equals to the square root of 25 meter, multiply to 9.8 meter per second squared, multiply to tan 20 degree. The final answer or the maximum speed of V is equals to 9.44 meter per second. Problem number 2. A car is to make a turn with the radius curvature of 60 meter at the speed of 10 meter per second. At what angle should the road be bank? Given the values of radius is equal to 60 meter with the speed of 10 meter per second, and the constant acceleration due to gravity is equal to 9.8 meter per second squared. The required on this problem is the angle of the road be banked or data. Given the formula of tan j and theta is equal to V squared over RG, and equate the formula to theta is equals to tan j and V squared over RG. Substitute the given values data in verse tan j and 10 meter per second squared over to 60 meter, multiply to 9.8 meter per second squared, cancel the meter and the second squared. The final answer or the angle of the road be bank is equals to 9.65 degree. 4. Trucks 6. Fit 5. Available