function that describes speed of object that accelerates and decelartes exponentially I am developing a program that requires math a little to hard for me. There is an object, let's call it car, moving between 2 points, A and B,  whose speed initially accelerates exponentially and then start to decelerate(symmetrically) to stop exactly at point B. The car takes the shortest possible path. I attached a photo to visually explain. 
Every 10ms I will have an opportunity to determine the next point that car should be at (basically I am in a run loop). How far away that point is from the current position, is determined by the speed the car currently has. The input to the program are just points A,B and time in which it has to perform the travel. 
I am struggling, how can I implement this behaviour within a run loop? I imagine I could do it if I knew what function to use for describing exponential acc- and deceleration shown in the graph.

 A: A function that comes to mind is the normal distribution as seen here.
You specified exponentials, however, your graph shows a smooth curve. Exponentials are involved in the normal distribution, but you can also create a cusp solution using two simple exponentials, $e^{t}$ and $e^{-t+offset}$ and a conditional branch.
There's an integral of the normal distribution called the error function which will be our main usage.
First you find out your distance from A to B, let's call it $\vec{v_0} = \vec{p_b} - \vec{p_a}$.
You then multiply the error function by the distance vector:
$$ \vec{v} = \vec{v_0} \cdot \bar{erf}(t) $$
where $t$ is time. You'll need to scale time dependent on how long you want the travel time to take.
Choose a sufficiently distant value to the left (say, $t_0 = -10$) of the normal distribution and iterate forwards in time (to $-t_0$), using your $erf(t)$ to compute the position. You can use the $normal(t)$ function to compute the instantaneous speed at the specified point.
In the next iteration you compute:
$$ t_{n+1} = t_n + dt $$
$$ \vec{v_{n+1}} = \vec{v_0} \cdot \bar{erf}(t_{n+1}) $$
$$ \vec{p_{n+1}} = \vec{v_{n+1}} + \vec{p_a} $$
Which gives you the position at the given time.
In the last step, if you do not use time values larger than about three standard distributions, you may need to complete the movement using a discrete step:
$$ \vec{p_{m}} = \vec{p_b} $$
($p_m$ is the final position)
because the error function ideally never reaches 1.
Note that you can customize the error function by specifying the standard deviation and the mean value.
Note that that $\bar{erf}$ is normalized between 0 and 1 for the computed vector to be correct. The standard error function ranges from -1 to 1.
