(original question, see edits below for full context)
After much frustration, I have figured out a function which maps velocity during acceleration/deceleration for my project.
$$\text{velocity} =s+\frac{-x^2\cdot 0.5\cdot j}{100000}$$ \begin{alignat}{1} x & = \text{time} \\ s & = \text{start velocity} & = 100 \\ j & = \text{jerk factor} & = 1 \\ \end{alignat}
For this function, $y$ represents velocity. Here's how it graphs using the above numbers...
I need to capture the point at which velocity will reach zero, so I know how many milliseconds $x$ it will take to come to a full stop.
If I know how long 'stopping' takes, I can use it to decide how soon a vehicle can stop to arrive at a point.
Given a provided $s$ and $j$, how can I calculate when $y$ will be zero? How do I arrive at $4472.136$ given the above numbers?
I believe the answer to the question involves polynomial factoring, but I have no idea how to get rid of the $x$ for this formula. I am aware this 'collides with zero' multiple times, but I only need one of those numbers since it's $-/+$
Can anyone shed some light? It is much appreciated
Edit
I've tested the chosen answer, and I'm pleased to say it works perfectly!
Jan Eerland's answer correctly finds the intersection with zero. I'm not really sure how. Actually I needed to remove the velocity from the equation for it to work...
This answer very quickly (and elaborately) solved the posted math problem. I'm posting my graph here with hopes my addition helps future readers. Since it solved the raised math question, it's the correct answer in this context.
However, R. J. Mathar pointed out my math doesn't make sense... and he's correct. I notice the curve should be linear, because my example does not dictate how the ^2 responds over time.
Here's a new formula which I think is correct, at least, it works in conjunction with my test spreadsheet numbers...
$$velocity\ =s+\frac{-x\cdot 1000\cdot j}{100000}$$
Actually...
It seems I just found the intersection for my new linear curve...
If anyone is wondering, I'm okay with my implementation having a linear decrease in velocity. In my implementation it will not really be noticed - so you should take that into account when deciding if this solution works for YOUR project...
I hope it helps someone out there!
$$x\ =\left(\frac{100s}{j}\right)$$
$$\text{Velocity}=\text{s}-\frac{jx^2}{200000}\Longleftrightarrow x=\pm\frac{200\sqrt{5}\cdot\sqrt{\text{s}-\text{Velocity}}}{\sqrt{j}}$$
Assuming $j\ne0$.
So, we get:
$$x=\pm\frac{200\sqrt{5}\cdot\sqrt{100-0}}{\sqrt{1}}=\pm2000\sqrt{5}$$
But time has to be positive so it is $x=2000\sqrt{5}\approx4472.14$
$$\text{Velocity}=\text{s}-\frac{jx^2}{200000}\Longleftrightarrow$$ $$\text{Velocity}-\text{s}=-\frac{jx^2}{200000}\Longleftrightarrow$$ $$-\text{Velocity}+\text{s}=\frac{jx^2}{200000}\Longleftrightarrow$$ $$200000\left(-\text{Velocity}+\text{s}\right)=x^2\Longleftrightarrow$$ $$\frac{200000\left(-\text{Velocity}+\text{s}\right)}{j} = x^2 \Longleftrightarrow$$ $$\pm\sqrt{\frac{200000\left(-\text{Velocity}+\text{s}\right)}{j}}=x$$