Electron speed in resistors
Basically, it can be explained by "Mechanics".

The voltage source at the resistor's ends, create a electric filed, assume the filed is uniform, and their is no other force on the particle, then

$$
\vec{F_{e}}=q\vec{E}=m\vec{a}
$$

So, the speed of the charge must have some relations with the voltage drop on the resistor.

Actually, the speed of the charge particle travel in a conductor, is called "drift velocity", in that link, it gives the drift speed equation.



If a thin wire and a large wire, given same voltage source, same material, and their main difference is only they have different area of cross-section. If the current flow in them are same, because

$$
I_{avg}=nqv_{d}A
$$

Where \$v_{d}\$ is the drift speed, \$n\$ is the charge-carrier density, \$q\$ is the charge on the charge-carrier, \$A\$ is the area of cross-section of the conductor. So the drift speed in the thin wire must greater then the "large wire".