Assuming that "faster" means acceleration then they are quite evenly matched, with the earlier model having a technical advantage because of its lighter weight.
In looking at the factory service manual specs for the '03 and '07 400s I find the following:
- the '07 model engine produces 3.2% more torque, 26.8 vs. 23.6 lb-ft;[/*:m:2vq7k0fj]
- the '07 model overall drive ratio (crankshaft to rear axle) is 13.0:1, vs. the early model's 13.6:1 ratio;[/*:m:2vq7k0fj]
- the '07 rear tire is 5.47% larger diameter, 21.27" vs. 20.17";[/*:m:2vq7k0fj]
From this data we can calculate the maximum thrust each machines can apply to the ground, as:
[lb-ft at the crank] * [overall drive ratio] / [rear tire radius (in ft)]
Hahahahahaha. You must've needed something to do?
for the '07; 26.8 * 13.0 / (21.27 / 2 / 12)
= 393.12 pounds thrust;
for the '03; 23.6 * 13.6 / (20.17 / 2 / 12)
= 381.91 pounds thrust;
Nonetheless at the bottom line even though the '07's maximum thrust is 2.94% more than the '03; 393.12 vs. 381.91 pounds--the '07 model's curb weight is 7.94% more, 462 vs. 428 pounds.
In a real race rider weight and the CVT torque/ratio curves (which are quite tunable) would be much more significant factors than any inherent difference in either machines' basic design...