# A 1200 kg car goes over a rounded hill in the road. The radius of the curve of the hill is 145 m. The car is going 100 km/hr. What is the centripetal acceleration the car feels at the top of the hill? What is the normal force the car feels at the top of the hill? How fast does the car need to move to make a jump/lose contact with the road?

First some conversions to get the values into the right forms:

velocity = 100km/hr = 27.8m/s

What is the centripetal acceleration the car feels at the top of the hill?

`F = m(v^(2)/(r))`  so `F=1200((27.8^(2))/145)`   and F = 6385.7N  (it doesn't look like significant figures are being considered in this problem,...

First some conversions to get the values into the right forms:

velocity = 100km/hr = 27.8m/s

What is the centripetal acceleration the car feels at the top of the hill?

`F = m(v^(2)/(r))`  so `F=1200((27.8^(2))/145)`   and F = 6385.7N (it doesn't look like significant figures are being considered in this problem, or else the answer would be 6000N)

What is the normal force the car feels at the top of the hill?

The centripetal force Fc may be considered to be the sum of the rider's weight and normal force added as vectors. Therefore

F = mg -  n

mg - n = Fc

mg - n = m(v^2/r)

In this case we're subtracting n from mg because the mg vector points in the same direction that the Fc does.

mg -m(v^2/r) = n

1200(9.8) - 6385.7 = 5374.3N

How fast does the car need to go to lose contact?

I assume we're asking about the car's velocity at the instant it  reaches the top of the hill. In order to lose contact, the car's velocity needs to increase such that the normal force holding the car "down" is equal to or less than 0.

0 = 1200(9.8) - 1200(v^2/145)

11760 = 1200(v^2/145)

9.8 = v^2/145

1421 = v^2

V = 37.7

So the velocity must meet or exceed 37.7m/s  or 135.7km/hr.