A car travels at a constant speed around a circular track whose radius is 2.5 km. The car goes once around the track in 430 s. What is the magnitude of the centripetal acceleration of the car?

Answer 1

The centripetal acceleration is
a = V^2/R,

and the velocity V is given by
2*pi*R = V*P

where P is the period of 430 s.

The second equation can be used to eliminate the variable V, leaving you with

a = [2 pi R/P]^2/R = 4 pi^2 R/P^2