Max spped for a car to go around a curve track without skidding....confusing!!! here's my question, a car of mass 1000kg moves along the corner of a level road having a radius of curvature 35.0m. If the frictional force between the tyres and the road is 4kN, the maximum speed of the car without skidding at the corner is A. 4.0 m/s B. 8.8 m/s C. 11.8 m/s D. 140.0 m/s ok, i knew there's a formula for the max spped of a car go around the curve track without skidding is v=(rag/h)^1/2, but is it possible to find the height form centre of gravity of car in the above case? I've tried usin' v=(Fr/m)^1/2 which is derived from the centripetal force, F=(mv^2)/r, nonetheless, if we using v=(Fr/m)^1/2, is this could be the maximum spped of car instead of using v=(rag/h)^1/2????