How do electrons in metals manage to have zero acceleration in constant $E$ field (as in a DC circuit)? From Newton's second law, a charged particle driven by a constant electric field should move with a constant acceleration. But electrons in a circuit acquire a steady average velocity which gives rise to a constant steady DC current. Of course, the electrons scatter with phonons and other defects leading to energy dissipation. But how does it explain zero average acceleration? What is the mathematical model which can be used to understand this?
 A: The simplest mathematical model would be a Newton's equation with viscous friction:
$$m\ddot{x} = -\nu\dot{x} -eE,$$
which is written in terms of position and velocity as
$$\dot{x} = v, \dot{v} = -\frac{\nu v}{m} -\frac{eE}{m},$$
and has a stationary solution for velocity:
$$v = -\frac{eE}{\nu}.$$
A bit more realistic model is obtained by adding a deltta-correlated random force:
$$m\ddot{x} = -\nu\dot{x} -eE + f(t),$$
where $\langle f(t) f(t')\rangle = D\delta(t-t').$
The average velocity, $\langle v\rangle$, here is given by the same solution as before, however one can also calculate the thermal fluctuations $\langle v^2\rangle$, which in a metal actually have higher velocity than the drift velocity of an electric current.
It is worth mentioning here the famous Drude model, although it is less suitable for description with Newton's equations: in this models the electrons do accelerate with a constant acceleration during the time between collisions. This model gives the same answer as discussed above, where the viscosity coefficient is replaced by $\nu = \frac{m}{\tau}$, where $\tau$ is the average time between collisions.
A: As you have indicated, scattering is the anwer.
The simplest model that deals with scattering events in solids is the Drude model of conduction. It assumes that charge carriers (electrons or holes) undergo frequent collisions, with a mean time of $\tau$, the relaxation time, between collisions. Carriers accelerate under the influence of the E-field, and each collision "resets" the carrier velocity back to zero, on average.
The average drift velocity over many carriers and through many collision events is
$$v_d=\frac{F\tau}{m}=\frac{qE\tau}{m}$$
where $q$ is the elementary charge and $m$ is the effective carrier mass. The mean distance between scattering events is called the mean free path, and given by
$$\lambda = v_{th}\tau=\sqrt{\frac{3k_BT}{m}}{\tau},$$
where $k_B$ is the Boltzmann constant and $T$ is the temperature. As long as the device dimensions are much greater than $\lambda$, the carriers can usually be safely regarded as moving with a constant drift velocity. However, in devices that are extremely short or have a very low scattering rate, carriers can traverse the device with few or no scattering events, accelerating the entire time. This is called ballistic transport.
A simple analogy for the finite drift velocity of carriers is objects reaching terminal velocity as they fall in atmosphere (see Vadim's answer). The weight of the object is analogous to the electric field: the greater it is, the higher (yet still finite) the terminal velocity.
A: In a nutshell:
Very few (if not none) electrons are accelerated all the way (no interaction with obstacles). Very few (if not none) will remain stationary (on every interaction they are sent back to where they started). But most of them (if not all) will follow the golden midway. They travel with constant (mean) velocity (acceleration, interaction, stop, acceleration, interaction, stop, etc).
A: When you are riding a bike in the atmosphere you put in a constant force yet you do not accelerate. Electrons have the same experience. The drag force is proportional to the velocity and therefore increases with velocity until it cancels the electric force. At this point the velocity remains constant.
