How to calculate the speed of a commuting bicyclist
Equations
Cyclist reach a steady speed when the motive power that they produce balances the rolling and air friction; in other words when the friction drags balance the motive power. 
The air drag is given by the well known equation:
Here A is the frontal cross section area, Cd is the drag coefficient, ris the density of air, The air drag is proportional to the velocity v squared. The dependence on the cross sectional area should make intuitive sense; its harder to shove a big object through the air than a thin object. The density comes into the air friction because the moving object has to shove the air out of the way.
The rolling drag is given by the equation:
Here M is the mass of the rider and bike, g is the acceleration of gravity, and Crr is the coefficient of rolling friction. This is simply the weight (mg) of the rider and bike times the coefficient of friction.
The total drag is the sum of these two drags:
The power required to overcome these drags is the velocity times the drag:
Thus the power required to overcome the drag scales as the velocity cubed.
To calculate some actual values, we need to know the values of constants. The density of air and the acceleration of gravity are well known.
The rolling coefficient is more obscure, but various sources give it as approximately
Certainly one can due better than this with a tuned racing bicycle, but the average commuter bike isn't in great shape. Fortunately the exact value doesn't matter much because the air drag is much more important than the rolling drag at high velocities.
The drag coefficient is commonly take to be 0.9
The cross sectional area is more problematic. Most of the measurements have been taken for racing cyclists, not commuters. Typical values quoted are 0.4 to 0.6m^2. Since a commuting cyclist rarely crouches, and is probably more upright than a racer using the top of her handlebars, I will use 0.67m^2. 
Finally, we need the mass of the cyclist and bicycle. Take a 150lb cyclist, with a heavyish bike of 28lbs.
Velocity vs. Power curves
Using these equations, we can easily calculate the speed of a cyclist as a function of the power the cyclist is putting out. For example, if the cyclist puts out 100w 
Here are two graphs of the power that a cyclist has to put out as a function of the cyclist's velocity.
Thus much over 20mph (10m/s) is very hard.
There are many calculators on the web that will tell you the power for any given velocity. Be aware that they do not always state the constants that they are employing. Two such calculators are at:
http://www.me.psu.edu/lamancusa/ProdDiss/Bicycle/bikecalc.htm
http://www.exploratorium.edu/cycling/aerodynamics1.html
A very good shareware program to calculate the power, (and many other things) is at
http://www.xsystems.co.uk/machinehead/
Stop signs and bicyclists
Frequent stop signs dramatically decrease the average speed of bicyclists. For instance, it turns out that the to maintain as speed of 12.7mph, the cyclist would have to increase her power output from 100 to 500watts on a street that has a stop sign every 300ft. Calculating this exactly is messy, so heres a simplified version:
Defining the distance between stop signs
The time between stop signs is
So every 16.1 seconds she would have generate, and then lose, all her kinetic energy. Now since she slows down near every stop sign, her peak speed must be higher than her average speed. Lets say she can accelerate and deaccelerate at a maximum rate of 0.15g. For comparison, a car going from zero to sixty in 13 seconds has an acceleration of 0.2g
Acceleration 
The approximate peak velocity can then be found by solving the equation:
The solution is
Her kinetic energy at this speed would be
She has to recreate this energy every 16 seconds, so her average power for this alone must be about
At this speed, the power necessary to compensate for the drag would be
Thus the total power she would have to supply would be
Note that this is very approximate. In particular, she is not at her peak speed all the time, do the drag power would be lower. Nonetheless, approximately the same answer is obtained when the calculation is done more exactly.