:EDIT:
Let me put some numbers on this.
Let's say you want to get from any angle to vertical in half a second. Say for now (more on this unrealistic scenario later) that you want to get from "laying down" (90 degrees from vertical) to upright in half a second. This is just for the purposes of coming up with a spec.
The position equation:
$$
\theta = \theta_0 + \dot{\theta}_0 t + \frac{1}{2}\ddot{\theta}t^2 \\
$$
(It's equally valid for linear or angular positions/speeds). You would think that you need to accelerate from 90 to 0 degrees, but then you'd hit 0 with some velocity. In reality, you're looking to get to 45 degrees in a quarter second, then spend the other quarter second decelerating. So, if you want a quarter second to go from $\theta = 90^{\circ}$ to $\theta = 45^{\circ}$, then:
$$
45^{\circ} = 90^{\circ} + (0^{\circ}/s)t + \frac{1}{2}\ddot{\theta}(0.25 \mbox{s})^2 \\
$$
$$
\frac{-45^{\circ}}{(0.25\mbox{s})^2} = \ddot{\theta} \\
$$
$$
\ddot{\theta} = 720 ^{\circ}/\mbox{s}^2 \\
$$
This is the acceleration you need beyond gravity to get the performance you want. You could accelerate slower if you want to - again, it's up to you to specify.
Check the top speed. At that acceleration, 720 degrees per second for a quarter second, you go from $\dot{\theta} = 0$ to $\dot{\theta} = (720^{\circ}/\mbox{s})(0.25\mbox{s})$, or $\dot{\theta} = 180^{\circ}/\mbox{s}$.
Now let's lower the speed requirement. Say instead of having the pendulum "pop" up from laying down in half a second, you're okay with it taking two seconds. Do the same steps as above, but now when it accelerates for half the time, it's accelerating for 1 second. It again decelerates for 1 second.
$$
45^{\circ} = 90^{\circ} + (0^{\circ}/s)t + \frac{1}{2}\ddot{\theta}(1 \mbox{s})^2 \\
$$
$$
\frac{-45^{\circ}}{(1\mbox{s})^2} = \ddot{\theta} \\
$$
$$
\ddot{\theta} = 45 ^\circ/\mbox{s}^2 \\
$$
Now you run at that acceleration for a full second, and get $\dot{\theta} = 45^{\circ}/\mbox{s}$.
Seems reasonable so far, but you have to torque the pendulum by accelerating the base. Find the projection of the arm to the +y-axis, using the angle the arm makes with the vertical ($\theta$), to see how the base acceleration impacts the and then see that the base generates an angular acceleration at $\ddot{\theta} = \frac{1}{\ell} \cos(\theta) \ddot{x}$.
Now, at 90 degrees, $\cos(\theta) = 0$, so the angular acceleration you get is zero for any $\ddot{x}$. No amount of base acceleration will get the pendulum upright.
Say though that you want to start at 89 degrees instead. Now $\cos(89^{\circ}) = 0.0175$, so
$$
\ddot{x} = \frac{\ell}{0.0175} \ddot{\theta} \\
$$
Or
$$
\ddot{x} = (53.3)(\ell)(\ddot{\theta}) \\
$$
So if $\ell$, the distance from the cart to the center of mass were 1 meter, then the cart has to accelerate 53 times the angular acceleration number. 720 degrees per second squared is 12.5 rad/s2, so you need your cart to accelerate about $(1\mbox{m})(53*12.5 \mbox{rad/s}^2) = 662.5 \mbox{m/s}^2$. At a gravitational constant of $9.81 \mbox{m/s}^2$, you're looking at a linear acceleration of about 67g.
For the slower case, 45 deg/s2 = 0.7854 rad/2, so now the cart would need to accelerate at $(1\mbox{m})(53*0.7854 \mbox{rad/s}^2) = 41.6 \mbox{m/s}^2$, or about 4.25g.
The faster case is accelerating at 67g for a quarter second, so $x_f = x_0 + v_0t + 1/2 at^2$, or $x_f = (0.5)(67\mbox{g})(9.81\mbox{m/s}^2/\mbox{g})(0.25\mbox{s}^2) = 20.54m$, versus the slower acceleration's case of $x_f = (4.25)(9.81)(1^2) = 20.84m$.
For top speeds, $(67)(9.81)(0.25) = 164 \mbox{m/s}$, versus the slower acceleration's $(4.25)(9.81)(1) = 41.7\mbox{m/s}$.
So my gut reaction earlier, that the slower acceleration would result in a higher top speed, was wrong. It can result in a longer distance traveled, but it looks like the top speed is lower.
But, I hope you can see that "popping up" from laying down is not realistic. You'd be better off to have some kind of spring-loaded trigger that pushed up instead of accelerating the base.
This still doesn't take into account the mass or moment of inertia; it's just looking at the base acceleration's relationship to a massless rod.
