Motion is a very diffuse concept :) you have to add a frame of reference to make it meaningfull.
In the frame of reference of the surrounding water the force definitely tries to stop the particle.
So if you have a stone rolled along the ground by a swift stream, the force goes in the direction of motion (in the usual, external, frame of reference), since the stone is still too slow for the water; whereas for a stone falling into a deep pond, the friction will be opposite ist motion.
a) The viscouse timescale is in play in the dynamics before reaching a steady-state situation.
Consider the equation of motion:
$$m\frac{dv}{dt}=\left(m-m_w\right)g - 6\pi\mu R_p v$$
Clearly, a steady-state is reached once the derivative vanishes, this occurs as $v\rightarrow v_p$:
$$0=\left(m-m_w\right)g - 6\pi\mu R_p v_p$$
Subtracting both equations yields:
$$m\frac{dv}{dt}= - 6\pi\mu R_p \left(v-v_p\right)$$
which can be rearranged using $m=\frac{4}{3}\pi \rho R_p^3$ to:
$$\frac{dv}{dt}= - \frac{9}{2}\pi\left(\frac{v-v_p}{\tau_v}\right)$$
where $\tau_v=R_p^2/\nu$ with $\nu=\mu/\rho$ which is the viscous timescale.
This equation dictatest that the velocity changes exponentially with a time-scale $\tau_v$ until a steady-state is reached ($v\rightarrow v_p$) because at that point $\frac{dv}{dt}\rightarrow 0$ and the rhs of the equation becomes $v-v_p \rightarrow 0$. As you see no dependence on the viscous time-scale once steady-state is reached.
So what other timescale is there to consider? Well the particle is now moving at a steady velocity, settling over a distance of height $H$ (i assume), so the equation of motion is now simply:
$$\frac{dv}{dt}=\frac{d^2y}{dt^2}=0$$
In other words:
$$ \frac{dy}{dt} = \frac{H}{t_s} = v_p $$
where $t_s$ is the settling timescale, i.e. $t_s=v_p/H$
b) As long as $\mathrm{Re}<1$, Stokes' flow is valid for non-uniform velocities however other effects may play a role.
Say that the non-uniform velocity is characterized by a shear rate of $\dot{\gamma}=\frac{U}{R_p}$ where $U$ is some characteristic velocity scale, then:
$$\mathrm{Re} = \frac{UR_p}{\nu} = \frac{U/R_p}{\nu/R_p^2}$$
For $\mathrm{Re}<1$ it requires that $\dot{\gamma}<1$, i.e. the velocity change over the particle will be small (maybe negligible).
If the velocity change over the particle is small then you may average the velocity over the particle (or maybe several particle radii) and use that as a value for the far-field uniform velocity, $v_\infty$.
You haven't specified exactly how the velocity field is in relation to the motion of the particle, either it is normal or parallel to the motion:
- normal: in this case no changes to the equation of motion are necessary, the normal velocity field serves to displace the particle in the horizontal direction. In this case you can use regular trajectory calculations to find where it lands.
- parallel: in this case a modification to the Stokes' equation needs to be made:
$$m\frac{dv}{dt}=\left(m-m_w\right)g - 6\pi\mu R_p (v \pm v_\infty)$$
As you can see the far-field velocity makes a contribution to the drag force:
- if it is in the direction of the motion of the particle, relative velocity is smaller, drag force is reduced so it is a negative contribution
- if it is against the motion of the particle, relative velocity is greater, drag force is increased so it is a positive contribution.
As mentioned at the start other phenomena start to play a role, specifically 'lift' becomes dominant. First theoretically treated by Saffman, 'lift' occurs when the particle will start a rotation due to the gradient in velocity and causes it to move laterally depending on the orientation of the flow in relation to the motion of the particle.
I will not go into details here, for more information see e.g. this publication on Saffman lift phenomena, specifically ch. 2 'Lateral migration: lift in low-Reynolds-number flows'
Best Answer
If you are modeling the surrounding fluid as a continuum (which you seem to be), then you can determine the drag force on the particle by integrating the tractions on the particle surface over the surface itself, or equivalently integrating the divergence of the fluid's stress field inside of the particle domain by virtue of Gauss's theorem;
$$\vec{F}_{drag} = \oint_{\partial \Omega} \bar{\bar{\sigma}} \cdot d\vec{a} = \int_{\Omega} \nabla \cdot \bar{\bar{\sigma}}\ dV$$
where the fluid stress tensor $\bar{\bar{\sigma}}$ is a constitutive function of $\nabla \vec{v}$ and other relevant variables in your problem (viscosity $\mu$, temperature $T$, etc.)
For a Newtonian fluid, the standard constitutive relationship applies, but if you're working with something a little bit more exotic like blood (which I'm betting could be the case since you're doing DEP), that relationship is different and your drag will be different than what you may expect from, say, Stokes's law.
In short, find what the flow field of your fluid is (with the particle in it), calculate the stress field using the appropriate constitutive relation for your fluid, and then calculate the integral above to obtain the drag force. You can do a sanity check by making sure your result is in the same ballpark as Stokes's law, which it should be barring the situation where you are working with a truly pathological fluid.