Angular Acceleration of a bicycle

[maniacp08]

A bicycle has wheels of 0.8 m diameter. The bicyclist accelerates from rest with constant acceleration to 22 km/h in 13.7 s. What is the angular acceleration of the wheels?

relevant equations:
omega = omega initial + alpha * t

22km/h = 6.11m/s

I did 6.11m/s * 13.7s = 83.7207 to get the velocity
then I divided it by .4m to get 209.30175
Im not sure what I should do nextI have another question which is related to the same topic:
A block of 2kg falls with a speed of 3.9m/s from rest to a distance of 2.5m, its acc. is 3.0m/s from a pulley of radius of 8cm.

(b) What is the angular velocity of the pulley at this time?
I used V = R/omega
= 3.9m/s = .08m / omega
omega = .02
This is wrong is my approach wrong?

 

[LowlyPion]

A bicycle has wheels of 0.8 m diameter. The bicyclist accelerates from rest with constant acceleration to 22 km/h in 13.7 s. What is the angular acceleration of the wheels?

relevant equations:
omega = omega initial + alpha * t

I did 22km/hr * 13.7s = 301.4 km s / hr to get the velocity
Vi = 0
Im not sure what I should do next

First convert km/h to m/s = 1000/3600

v = 22 kmph = 6.1 m/s

w = v/r = 6.1/.4 = 15.278 radians

w = a*t => a = w/t = ... ?