Question

The car travels at a constant speed from the bottom A of the dip to the top B of the hump.if the radius of curvature of the road at A is PA=120 m and the car acceleration at A is 0.4g, determine the car speed v.if the acceleration at B must be limited to 0.25g, determine the minimum radius of curvature pB of the road at B.

Answer 1

Since $v=\sqrt{a\times r}$ , then $v=\sqrt{0.4\times10\times120}\approx6.93$ m/s;

Since the speed is the same, the radius is equal to: $r=\frac{v^2}{a}=\frac{6.93^2}{0.25\times10}=19.20996$ m