How to find the maximum velocity a particle can have to stop within a given distance I have a set of particles moving with different velocities which are being decelerated by a laser. I have calculated the distances travelled by solving the differential $$dv/dt = F(v)/m.$$ The particles need to stop within a given length of the trap, so I then subtracted the values of the particle's travel distance from the length of the trap to get the particle's stopping distances. My question is: how I do use this information to then find the maximum velocity a given particle can have to stop within the trap?
 A: start with
$$m\,\frac{d^2 x}{dt^2}=F(v)=F\left(v\mapsto \frac{dx}{dt}\right)$$
you obtain the solution $~x(t)~$
solve $~x(t)=L~$ for t, you obtain $~t_L~$
thus the velocity at $~t_L~$ is
$$ v=\frac{dx}{dt}\bigg|_{t=t_L}$$
A: If acceleration (or deceleration) is a function of velocity only, such that $$ a(v) = \tfrac{1}{m} F(v)$$
then the distance traveled to stop from the initial velocity $v_0$ is
$$ \Delta x = \int \frac{v}{a}\,{\rm d}v = \int _{v_0} ^ 0 \frac{m v}{F}\,{\rm d}v $$
actually, since $F(v)$ is always negative above, you rewrite the above with a positive $F(v)$ by flipping the limits of integration.
$$ \Delta x = m \int ^{v_0} _ 0 \frac{v}{F}\,{\rm d}v $$
To find the maximum $v_0$, you have to solve the above for $v_0$. This will give you the velocity to reach the trap at $\Delta x$ given the forcing function $F(v)$.
But if you want, take the forcing function and parameterized it, like in $F(v) = \alpha - \beta v^2$ and then tune the parameters to maximize $v_0$ by taking derivatives
$$ \begin{aligned} \frac{\partial v_0}{\partial \alpha} & = 0 &  \frac{\partial v_0}{\partial \beta} & = 0  \end{aligned}$$
and the same for the remaining parameters (if they are independent of each other).
In the above example $\Delta x = \frac{m}{2 \beta} \ln \left( \frac{\alpha-\beta v_0^2}{\alpha} \right)$ which is solved for $v_0$.

The the derivative $ \frac{\partial v_0}{\partial \beta}=0$ yields the following optimization expression
$$ \frac{ \sqrt{\alpha} \left( e^{ \frac{2 \beta \Delta x}{ m}} ( 2 \beta \Delta x - m) + m \right)}{2m (-\beta)^{3/2} \sqrt{e^{\frac{2 \beta \Delta x}{m}}-1}} = 0 $$
which can be used to find the optimal $\beta$ to maximize $v_0$.

I realized the above yields a solution of $\alpha =0$ and $\beta =0$ which is not a solution. I need the real forcing function to really try to find the optimal solution.
