Question

The tires of a car make 61 revolutions as the car reduces its speed uniformly from 93.0 km/h to 63.0 km/h. The tires have a diameter of 0.80 m.

Part A What was the angular acceleration of the tires?

Part B If the car continues to decelerate at this rate, how much more time is required for it to stop?

Part C If the car continues to decelerate at this rate, how far does it go? Find the total distance.

Answer 1

Vinitial = 93/3.6 = 25.833 m/sec
Vfinal = 63/3.6 = 17.5 m/sec

Distance = Revolutions*pi*D = 61*0.98*3.14 = 172.47 m

?t = 2d/(Vf+Vi) = 8 sec
a = ?V/?t = (25.833-17.5)/8 = 1.04 m/sec^2

Part A What was the angular acceleration of the tires?
? = a/r = 1.04/0.4 = 2.6 rad/sec^2

Part B If the car continues to decelerate at this rate, how much more time is required for it to stop?
ts = Vf/a = 17.5/1.04 = 16.82 sec

Part C If the car continues to decelerate at this rate, how far does it go? Find the total distance.
slowing space = Revolutions*pi*D = 61*0.98*3.14 = 172.47 m
total stopping space = 1/2*Vi*(?t + ts) = 0.5*25.833 * (8+16.82) = 320.587m