I give you the result of my calculus without the details:
The result is :
$$ \Delta t = \frac{\Delta x}{V_{Max}} + \frac{1}{2}(\frac{V_{Max}}{a} + \frac{a}{j}) + + \frac{1}{2}(\frac{V_{Max}}{d} + \frac{d}{j})$$
where :
$\Delta t$ is the total time.
$\Delta x$ is the total displacement.
$a$ is the maximum acceleration.
$d$ is the maximum decceleration.
$j$ is the jerk.
$V_{max}$ is the maximum speed.
As a test, with your values :
$\Delta x = 2 m, a = 1m/s^2, d = 1m/s^2, j = 1 m/s^3, V_{max} = 1m/s$, I find :
$$ \Delta t = \frac{2}{1} + \frac{1}{2}(\frac{1}{1} + \frac{1}{1}) + + \frac{1}{2}(\frac{1}{1} + \frac{1}{1}) = 2 + 1 + 1 = 4$$
which is the correct result.
So I am quite confident in the formula.
[EDIT]
The formula to obtain jerk is :
$$j = \frac{a + d}{2 (\Delta t - \large \frac{\Delta x}{\large V_{max}}) - V_{max}(\large \frac{1}{a} + \large\frac{1}{d})}$$
[EDIT 2]
The used model is :
Phase 1 : constant (positive) jerk $j$
Phase 2 : constant acceleration $a$
Phase 3 : constant (negative) jerk ($- j$)
Phase 4 : constant speed $V_{Max}$
Phase 5 : constant (negative) jerk ($- j$)
Phase 6 : constant decceleration ($d$)
Phase 7 : constant (positive) jerk ($j$)
In the formulas above, there are constraints, more precisely the duration of the phases 2, 4, 6 must be positive:
$$\Delta t_2 = \frac{V_{Max}}{a} - \frac{a}{j}\ge 0$$
$$\Delta t_4 = \frac{\Delta x}{V_{Max}} -\frac{1}{2}(\frac{V_{Max}}{a} + \frac{a}{j}) - \frac{1}{2}(\frac{V_{Max}}{d} + \frac{d}{j}) \ge 0$$
$$\Delta t_6 = \frac{V_{Max}}{d} - \frac{d}{j}\ge 0$$
If one of these constraints is not satisfyed, this means that the hypothesis taken for the model are incoherent, so we need another model.
Best Answer
Your robot surely needs this...
Consider your robot (Let's name him Rob!!) has been destined to have a maximum acceleration $\alpha$ and maximum deceleration $\beta$. For this time interval $t$, Rob covers a distance of $s$. Let's work this from a graphical approach rather than using derivatives :
Now Rob's maximum velocity $v_m$ is represented by the peak Q.
All inferences has been based on the graph.
$T=t_1 + t_2$
Slope of OQ curve= $\alpha$=$\frac {v_m}{t_1}$
Slope of PQ curve=$\beta$=$\frac {v_m}{t_2}$
From here let's put the values of $t_1$ and $t_2$ in the first equation which goes by-
$$T=\frac {v_m}{\alpha} + \frac {v_m}{\beta}$$
$$\Rightarrow v_m=\frac {\alpha \beta T}{\alpha + \beta}$$
For Rob's $v_m$ to be a function of $s$ you can easily do by using other physical consequences of the graph that I leave upto you.
Now Rob looks polished.
Cheers!!