# A car drives over a hill with a circular top that has a radius of curvature of 10m.How fast can it go at the top of the hill if it is not to leave the road?

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- We know when the car is traveling on top of a curve, a centripital force (`mv^2/r` )will act on the car which will result in leaving the car from the road.
- This force in increasing with the increase of velocity(`v` ) of the car.
- Due to the friction normal force will act on the car at same direction of the centripital force.
- Both normal and centripital force was balanced by the weight(`mg` ) of the car.
- When the addition of centripital force and normal force exceed the weight of the car, the car will leave out from the road surface and lift up.
- The maximum speed which can achive without leaving the car is the speed that centripital force = weight of car
- In this maximum speed the car is about to leave and normal force can be considered as 0.

Centripital force = m*v^2/r

Weight of car = m*g

mg = m*v^2/r

v= sqrt(gr)

= sqrt(9.81*10)

= 9.9 m/s

**So the velocity which can travel the car without leaving the road is 9.9 m/s**

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