Less pollution: moving hurricane debris to other regions for use, or burning? When a big hurricane hits, it can create debris on the scale of $\mathrm{10^8 yd^3}$.  Cities in Florida, Texas, and other affected areas are struggling to hire enough trucks and drivers to pick it up quickly.  But aside from that, I noticed many of the areas have started to burn the debris once it starts building up.

Got to wondering... typically mulch comes in modestly pricey, and when free mulch is offered, it often goes quickly.

So assuming a fair portion of debris is mulchable and is of interest to other areas, and that we can acquire typical transportation resources, then we'll set up transfer from collection sites to those other regions rather than burning it.  What would be the net pollution result?

If removed for mulch and such: trucking pollution + decomposition (- trees saved locally??)
If burned: the burning pollution.

Obviously it's about approximation rather than exacts, it's probably hard to appraise the different byproducts from burning versus decomposition, and a lot probably depends upon the way it is burned.  But as a whole, can we get a rough estimate of comparable quantities/damage done... is it less pollution/damage even to truck it an average of 3000 miles?  1000 miles?  100 miles?  10 miles?  Should it be burned on the spot (if done safely)?  Would think there's got to be some way to get a very rough idea.

Certainly the best option if viable might be leaving it in place to decompose.  But considering how upset people are getting at having debris around these parts a month later, exclude that option from the possibilities.

Trucking or burning, how do they compare?
As the question was changed, my answer attempts to evaluate only the difference between burning and transporting. Please correct my values if my quickly found sources are inaccurate or you find more representing. I know there is quite a few unwritten assumptions that simplify this problem.


On average transporting via trucks emit 161.8 $\frac{CO_{2}}{short-ton-mile}$ (according to this Environmental defense fund handbook). Short ton equals to 2 000 lbs.
The amount of $CO_{2}$ equivalent from burning wood is $1900 + 200 + 70 = 2170 \frac{CO_{2} eqiuv.}{kg}$ from carbon dioxide, nitrous oxide and methane respectively (according to this article, the link to the source of these numbers fails so please comment if these numbers aren't correct).


Now we can write an equation out of this and solve for the kilometer distance that produces equal amount of carbon dioxide equivalent.$$\frac{2170\frac{CO_{2}eq.}{kg}}{161.8\frac{CO_2}{short-ton-mile} \cdot 0.0011\frac{short-ton}{kg} \cdot 0.6213\frac{mile}{km}} \approx 20000 km$$
This is quite a large number in my opinion. I would've expected less. So quite a long distance is needed until transporting produces more $CO_2$ pollution than burning.