5V Regulator Power Dissipation
The quick answer is:  You're probably over the temp/power limit for that part, on that PCB.

The not-so-quick-but-more-accurate answer is:  

Using the numbers from the datasheet, combined with some thermal conductivity numbers for your PCB, factoring in the ambient air temp and ventilation you should be able to calculate the junction temp of the regulator (and this, if you are exceeding the max temp).  

That sounds easy, but it is not for most engineers.  The problem is that you don't really know your airflow, or the thermal conductivity of the copper on your PCB.  You could make some guesses and get somewhat close, but in the end you'd have to just try it and see.  

Or, you can use experience and a rough guide.  This isn't super accurate either, but it gets you in the right ballpark.  My experience says that 1 cm^2 is not enough to dissipate 2 watts.  I would go with at least 4 cm^2 and 6 cm^2 would be much better.  But even if you get 6 cm^2, that doesn't mean you're safe.  If you are running with high ambient temperatures or don't have much airflow, or there are other hot components nearby then you might still have issues.

Most professional EE's in this situation would do as much math as possible, make an educated guess, then make the copper as large as possible.  Then, after the prototypes are built, verify the design by running it at max current, at max ambient temperature.  Only then would an EE be satisfied that all the requirements have been met.

If this were my design, I would use a switching regulator instead of a linear regulator.  It will be smaller, and waste less power.  But it is more expensive and more complicated.