Answer:
34.3 m/s
Explanation:
Newton's Second Law states that the resultant of the forces acting on the car is equal to the product between the mass of the car, m, and the centripetal acceleration [tex]a_c[/tex] (because the car is moving of circular motion). So at the top of the hill the equation of the forces is:
[tex]mg-R = m a_c = m\frac{v^2}{r}[/tex]
where
(mg) is the weight of the car (downward), with m being the car's mass and g=9.8 m/s^2 is the acceleration due to gravity
R is the normal reaction exerted by the road on the car (upward, so with negative sign)
v is the speed of the car
r = 0.120 km = 120 m is the radius of the curve
The problem is asking for the speed that the car would have when it tires just barely lose contact with the road: this means requiring that the normal reaction is zero, R=0. Substituting into the equation and solving for v, we find:
[tex]v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(120 m)}=34.3 m/s[/tex]
