Question

A car travels at a constant speed around a circular trackwhose

radius is 2.6 km. The goes once around the trakc in 360s. What is
the magnitude of the centripetal acceleration ofthe car?

Answer 1

a = 0.77 m/s^2

Step-by-step explanation:

The centripetal acceleration is given as

`a = (v^2)/(R)`

The velocity of the car can be calculated using the circumference of the track and the period of the car.

`2\pi R = v T\\2\pi 2.6 = v 360\\v = 0.045 ~km/s = 45~m/s`

So, the acceleration is

`a = (45^2)/(2.6* 10^3) = 0.77~m/s^2`