I have $2$ moving objects (see picture) both with the same size and acceleration (which is negative because of friction but I'll keep it positive and change the formula instead) $$2R = 1, a = 1$$ I need to find time t when these 2 objects collide with each other knowing:
- initial position $(x,y)$
- initial speed $v^2 = v_x^2+v_y^2$ (the orange arrow, different value for each object: $v_1, v_2$)
and based on this, each object can travel a specific distance $d_m=\frac{v^2}{2a}$ (with green) but at some point they might collide with each other. That means: $$(2R)^2 = \Delta d_x^2+\Delta d_y^2$$ Where $(d_x,d_y)$ is the position of an object at impact $$d_x = x + d*\frac{v_x}{v}$$ and from physics: $d = t * (v - t*\frac{a}{2})$
and $d^2 = (d_x-x)^2 + (d_y-y)^2$ but this one might not be needed.
results in: $d_x = x + t * (v - t*\frac{a}{2})*\frac{v_x}{v}$
After all these I end up with a very long formula which simplified looks like this
$$(2R)^2 = (c_x+b_x*t-a_x*t^2)^2 + (c_y+b_y*t-a_y*t^2)^2$$
where for e.g. $\Delta d_x = (x_2 - x_1) + (v_{2x} - v_{1x})*t - \frac{a}{2} * (\frac{v_{2x}}{v_2} + \frac{v_{1x}}{v_1}) * t^2$
so: $c_x = (x_2-x_1)$
Apparently this is a $4$th degree equation but we know that $2$ objects cannot collide more than once so there must be a much easier solution or some special case.
I probably picked the wrong formulas or made some bad assumptions.
I need to find the most simplified formula for t
Picture with initial state
Your equation is correct, the time where the objects collide is the smallest solution of the equation. Unless this is a special case, that equation can't be simplified any better since if you're able to solve this, you would also be able to find the smallest root of those not very special polynomials. Also, the equation can be derived more neatly if you choose an object as frame of reference.
If you think about it, a 4th degree polynomial is not unexpected because of the functions that describe the position of the objects doesn't just stop after they collide. According to the function, they would just pass through each other and "touch" again while moving away. Due to negative acceleration, they would go backward and possibly "touch" two more times.