A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill.At what speed would the car be traveling when it tires just barely lose contact with the road when the car is at the top of the hill?
A) 45.5 m/s
B) 41.8 m/s
C) 34.3 m/s
D) 22.2 m/s
E) 27.6 m/s

Question

Quizplus · Accepted Answer

For the tires to just barely lose contact with the road at the top of the hill, the centripetal force must equal the gravitational force acting on the car. The formula for centripetal force is $F_c = \frac{mv^2}{r}$, where $m$ is the mass of the car, $v$ is the velocity, and $r$ is the radius of curvature. The gravitational force is $F_g = mg$, where $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$). Setting these two forces equal gives $\frac{mv^2}{r} = mg$, which simplifies to $v^2 = rg$. Plugging in $r = 0.120 km = 120 m$ and $g = 9.8 m/s^2$, we find $v = \sqrt{120m \times 9.8 m/s^2} = \sqrt{1176} m/s \approx 34.3 m/s$.

A Car Drives Over a Hilltop That Has a Radius