What is the outer radius of a car tire that has tangential speed of 36m/s and a period of rotation of 5.60*10e-2?

If the period of rotation, T,  of the car tire is seconds, then the frequency of the rotation, f, can be found as

The angular speed of the tire, , relates to the frequency as

rev/s (revolutions per second.)

If the tangential speed of the tire is 36 m/s, it means that a point on the outside of the tire (a point on the rim of the tire) moves with the linear velocity of 36 m/s. This point has the radius of rotation equal to the outer radius of the tire, r.

The relationship between the linear velocity and the angular velocity is

Then, the outer radius of the tire is

.

The outer radius of the car tire is 0.32 meters, which is 32 centimeters, or approximately 1 ft.