I think Ced is correct here; parabolas are going to get you the most bang for your buck where minimizing the second derivative is concerned.
This answer gives a slightly more generalized solution for achieving that parabolic shape, and other shapes (in case your requirements change a bit, maybe). I'm going to make a few assumptions based on the image in your question:
Your input is something like a step function; that is, the same value appears several times in a row before a different value appears.
Your maximum allowed second derivative is large enough that your output signal can usually "catch up" to the input signal before that value changes.
You don't want any overshoot (ringing/clipping).
Parabolic shape
First, let's express the parabolic shape as a simple equation. We'll have it begin at $(-1,-1)$ and end at $(1,1)$; in other words we'll have it cover two distance units (Y) over two time units (X).
$$y = x\left(2-\left|x\right|\right)$$
It looks like this, along with its first and second derivatives. From now on I'll refer to the first derivative as "velocity" and the second as "acceleration."

The maximum acceleration here is $2$; call that $A_0$. The distance along Y is also $2$; call that $D_0$.
We can scale this thing along X and Y to get a maximum acceleration of your choosing (call it $A_1$), over a desired distance (call it $D_1$). The X-scale and Y-scale can be calculated like this:
$$x_s = \sqrt{\frac{A_1D_0}{A_0D_1}}\\y_s = \frac{D_1}{D_0}$$
Now we'll modify that first equation to multiply all $x$ by $x_s$, and multiply the entire thing by $y_s$.
$$y = (x_sx)\left(2-\left|x_sx\right|\right)y_s$$
So, let's say you want the maximum acceleration to be $\frac{1}{4}$, and need to travel a distance of $1$.
$$x_s = \sqrt{\frac{A_1D_0}{A_0D_1}} = \sqrt{\frac{\frac{1}{4}\cdot2}{2\cdot1}} = \sqrt{\frac{1}{4}} = 0.5$$
$$y_s = \frac{D_1}{D_0} = \frac{1}{2} = 0.5$$

At this rate of acceleration, it takes four units of time (X axis) to travel one unit of distance (Y axis). You can calculate the time it will take like this:
$$t = 2\sqrt{\frac{A_0D_1}{A_1D_0}}$$
Other shapes
The nice thing about this is you can use pretty much any shape you want and those calculations will still work. Let's say you want to use a sinusoidal curve instead of parabolic (for aesthetic reasons, maybe).
$$y = \sin\left(\frac{\pi}{2}x\right)$$
Again, we have it begin at $(-1,-1)$ and end at $(1,1)$.

The maximum acceleration here (our $A_0$) is $\frac{\pi^2}{4}$, or $\approx 2.4674$ as we can see from the plot. Our $D_0$ is always $2$. You can calculate the X-scale, Y-scale and time just like we did for the parabola; it'll still work fine.
Or instead of the sine shape, try something like this:
$$y = \frac{15x-10x^3+3x^5}{8}$$
Acceleration is zero at each end; no jerky "takeoffs" or "landings."
(work in progress...)