As I pointed out much earlier, starting with reply #20 to this thread, you can pretty much explain it without much more than the principles of Archimedes.
Indeed, and for reference, here is an analysis of the cart/wheel/belt system.
Refer to the diagram attached below.

Assume that it is an ideal system, no friction losses, and both wheels are able to spin freely on their bearings. No slip occurs between the wheels and the belts.
There are five variables: \$v_W\$, \$v_R\$, \$v_M\$, \$v_G\$ and \$v\$
\$v_M\$ and \$v_G\$ are the rim speeds of the wheels, in m/s.
\$v_W\$ and \$v_R\$ are given, leaving three degrees of freedom.
Two equations can be written relating the belt velocities, the rim speeds of the wheels, and the cart speed:
$$v=v_M-v_W$$ $$v=v_G-v_R$$
These two equations take up two degrees of freedom, leaving one degree of freedom remaining.
With no additional constraints this system has an infinite number of solutions. The cart velocity, \$v\$ can have any value we choose.
This makes sense, intuitively, since the wheels are free running, and without friction losses any initial cart velocity will remain the same forever.
To remove the remaining degree of freedom in the system, the wheels can be linked with a connecting gear. If we let the gear ratio be \$\alpha\$, then we can write a linking equation:
$$v_G=\alpha\,v_M$$
For each revolution of wheel M, wheel G makes \$\alpha\$ revolutions.
(Note that the gearing could be mechanical, or it could be electrical. Mechanical is simpler.)
By making a suitable choice of \$\alpha\$, we can arrange for a desired cart speed, \$v\$.
For instance, suppose we wish the cart to move at 1 m/s to the right.
We set \$v\$ = 1 m/s, which gives \$v_M=v+v_W\$ = 5 m/s, and \$v_G=v+v_R\$ = 11 m/s.
From this, we have: \$\alpha = v_G / v_M\$ = 11 / 5 = 2.2
Therefore, with a gear ratio of 2.2, the cart will move to right at a speed of 1 m/s.