Drude model: A concerning approach to $\tau$ hurts Ohm's Law 
Here's the question. Between two points, $a$ and $b$, if an object is accelerated between them, it will cross the distance between them in time $t=\sqrt{2\frac{d}{a}}$. So, if the acceleration is increased, the time decreases with the square root of $a$.
If we take the average $t$ for sinusoidally distributed $d$ values, our $\tau$ should be $\tau = \frac{t}{\sqrt{2}}$. Therefore, increasing the $E$ field should decrease the value of $\tau$ by the square root of the increase. Therefore, the current density equation $j=nq^2\frac{\tau}{m}E$ should be scaling up $j$ by the square root of $E$ since we substitute $\tau=\sqrt{\frac{d}{E}}$. This defeats Ohm's Law. I don't want it to though because its seemingly wrong, but I don't know why.
 A: The electric field (or whatever force) has only a very small effect on the time between collisions.  The reason is that electrons are very light, and their thermal velocity exceeds any change in the velocity due to the applied fields.
One thus treats the scattering time (tau) as a constant.
The applied electric field then causes a drift velocity, tau*acceleration.
Oh, it's assumed that every collision resets the electron velocity randomly... it doesn't remember it's previous velocity.  
A: That seems right, but due to your unconventional (most of the time, like below, they do it with velocity, not distance) derivation of the electron kinematics (which are, I should point out, just an approximation), I think the source of your problem is that you still have $d$ in your final Ohm's law. If you replaced that by the more usual terms ($\tau$ and $E$), you'll probably get the right relation.
Here is the quick little derivation from p7 of Ashcroft and Mermin:
Common definition of current: $j = nqv_{avg}$
$t$ is time since electron's last collision. In this time it has force $qE$ acting on it, so accelerates at $a=qE/m$, so in time $t$ will gain velocity $v=at=qEt/m$. Plugging this into the above definition for current, $j=nqv=nq^2tE/m$.
(an important little detail they mention is that we assume the electron left its last collision with a random velocity $v_0$, but those all average out among all the electrons. The velocity from $E$ doesn't, though.)
