Relation between Angular speeds/accelerations

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I have been having trouble with the following question on my homework:

wheel A of radius rA = 15 cm is coupled by belt B to wheel C of radius rC = 22 cm. The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad/s2. Find the time needed for wheel C to reach an angular speed of 140 rev/min, assuming the belt does not slip.

HINT: The constant angular-acceleration equations apply. The linear speeds at the rims are equal. What then is the relation between the angular speeds and the angular accelerations?

please help, thank you for your time.
 
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For reference:

http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

Now there are two concepts here.

1. A disc starts from rest at constant angular acceleration.

So find the equation that gives the angular velocity at the surface of A, at constant angular acceleration at time = t.

2. Then determine the tangential velocity of the belt B, v(t).

The belt drives disc C, so one has to convert v(t) to the angular velocity of C and when it reaches 140 rpm (pay attention to units; angular velocity is in rad/s). It's somewhat the reverse process of part 1.
 
the only equation i can think of that relates to a constant acceleration is v1 = v2 +at

i need some more help because i know that isn't right...

i found the time to be .4375s
 
One needs the analog of v1 = v2 +at for rotational motion.

[tex]\omega(t) = \omega_0 + \alpha*t[/tex], where [itex]\omega(t)[/itex] is angular velocity, and [itex]\alpha[/itex] is angular acceleration.

The linear or tangential velocity at radius r is just v(t) = [itex]\omega(t)[/itex]*r,

and dividing by r, [itex]\omega(t)[/itex] = (v(t)/r.
 
alright, that helped a lot. now i see the relationship and substitution... also i was using A's radius, not C's.