The small cart will reach wind speed (stationary on a level treadmill) at 2.7 mph. At 10 mph treadmill speed, the 169 gram cart will remain stationary on a 4.4 degree incline, indicating a force imbalance in the forward direction of about 13 grams - enough to accelerate the cart at about 2.47 ft/sec2 if it was at 10 mph on a level road in a 10 mph wind. Once above the break-even speed, the cart will get closer and closer to it's theoretical advance ratio as the wind speed increases. It will always be above the wind speed once the wind speed gets past the break-even point.
Jack Goodman's cart has a break-even speed of 4 mph, and Mark C's cart (which spork and JB copied and refined) has a break-even speed of 8.5 mph. I don't know if Mark C's treadmill was level or inclined - can't remember.
In this article, Jack measured his cart's performance on the treadmill. He was also kind enough to send me the results of each 1 mph interval between 4 and 10 mph so that I could graph them.
http://www.ayrs.org/DWFTTW_from_Catalyst_N23_Jan_2006.pdf
His cart will easily start and roll down a one inch in ten foot incline. He measured the rolling resistance at 12.5 grams. At 4 mph the cart is at windspeed. At 10 mph it pulls forward with 150 grams of force. Each 1 mph step between 4 and 10 mph shows an increase in that unbalanced force of 25 grams with about 5% variation.
I suspect that there will be a range in which the cart will achieve the highest multiple of wind speed based on the efficiency of the drive train and propeller. I'm not sure about an ultimate top speed, because as the wind speed increases, the energy available also increases. The limiting factor would likely be the traction of the drive wheels.