Question

A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in 9.00 s. If the diameter of the tire is 58.0 cm, find (a) the number of revolutions the tire makes during the motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second?                                          

Answer:  (54.3 rev, 12.1 rev/s) --> please show me how to get this answer!

Answer 1

  22m/s from rest with constant acceleration means that the average speed was 11m/s.
Therefore, you can get the distance travelled by distance = time * average speed:
distance = 9s * 11m/s = 99m

Now, to get the number of revolutions, you have to divide the total distance by the circumference of the tire. The circumference is equal to pi * diameter:
Circumference = pi(3.14159265) * 0.58m = 1.82m

So:
Revolutions = 99m / 1.82m = 54.33 revolutions.

The angular speed can be worked out using the circumference as well.
22m/s is the final speed, so if you divide that by the circumference, you get the revolutions per second, which is your angular speed:
22m/s / 1.82m = 12.07 revolutions per second.

If they want this in degrees per second, just remember that there are 360 degrees in a revolution, so mutliply it by 360:
12.07 * 360 = 4364.58 degrees per second.

Final answers:
a) 54.33 revolutions
b) 12.07 revolutions per second or 4364.58 degrees per second