The motion model is defined by
Initial position $\vec{x}$ in 2d space
Initial timestamp $t$
Initial velocity $\vec{v}$ of $x,y$ direction
a constant acceleration $\vec{a}$ of $x,y$ direction
How can I find the timestamp $t$ that, two object $(\vec{x}_0, \vec{v}_0, \vec{a}_0, t_0)$ and $(\vec{x}_1, \vec{v}_1, \vec{a}_1, t_1)$ are/were closest to each other?
What I have try:
Let $p_0$ and $p_1$ be the point that they are/were closest to each other at some timestamp $t$ which can be calculated from $p_0=x_0+v_0(t_0-t)+\frac{1}{2}a_0(t_0-t)^2$
I tried to solve $t$ by throwing this into matlab/python sovler $\frac{d|p_0-p_1|_{2}}{dt}=0$. but it give me a huge mess and I am wondering if there is a simpler way to do it?
Edit Note: the two point might not collide
I think the formula you want for predicting positions is $$\vec p_0 = \vec x_0 + \vec v_0(t - t_0) + \tfrac12 \vec a_0(t - t_0)^2.$$ (The object tends to move in the direction of $\vec v_0$ as time increases.)
But you can simplify things by rewriting the conditions so the initial timestamp is the same. For example, you can make the initial timestamp $t_0$ in both cases by replacing $\vec x_1$ and $\vec v_1$: \begin{align} \vec x'_1 &= \vec x_1 + \vec v_1(t_0-t_1) + \tfrac12 \vec a_1(t_0-t_1)^2, \\ \vec v'_1 &= \vec v_1 + \vec a_1(t_0-t_1). \end{align}
So now you can describe the second body as $(\vec x'_1, \vec v'_1, \vec a_1, t_0).$
You can change the problem to one of relative position/velocity of the two objects. Let \begin{align} \vec x_2 &= \vec x'_1 - \vec x_0,\\ \vec v_2 &= \vec v'_1 - \vec v_0,\\ \vec a_2 &= \vec a_1 - \vec a_0. \end{align}
Then the relative position of the two objects at any time $t$ is $$\vec p_2(t) = \vec x_2 + \vec v_2(t - t_0) + \frac12 \vec a_2(t - t_0)^2.$$
You want to minimize the length of the vector $\vec p_2(t).$ This amounts to finding the point on a parabola closest to the point $(0,0).$ You have the point on the parabola as a function of time, the slope of the normal also obtainable as a function of time, and you need the line with that slope through that point to pass through $(0,0).$