#### Answer

a. Point C
b. Point A
c. Point C
d. Point A
e. Up to a velocity of $v = \sqrt{Rg}$

#### Work Step by Step

a. The normal force is largest at point C, the valley. The centripetal force is directed upward. The 2 forces acting on the car at that instant are gravity (pulling downward) and the normal force from the road (pushing upward). The net force is upward, so the magnitude of the normal force is greater than the car’s weight.
b. The normal force is smallest at point A, the hilltop. The centripetal force is directed downward. The 2 forces acting on the car at that instant are gravity (pulling downward) and the normal force from the road (pushing upward). The net force is downward, so the magnitude of the normal force is less than the car’s weight.
c, d. The driver’s apparent weight equals the normal force on her body, so she would feel heaviest at point C, and lightest at point A.
e. At point A, losing contact with the road means that the normal force is zero. At that critical speed, the centripetal force is just equal to the weight.
Find the critical speed by solving $mg = \frac{mv^{2}}{R}$ for v.
$$v = \sqrt{Rg}$$