= Solution
A <Newtonian fluid> exerts viscous <shear stress> proportional to the rate of strain. For an incompressible fluid of constant <dynamic viscosity> $\eta$,
$$
\boldsymbol\sigma=-p\mathbf I+\eta[\nabla\mathbf u+(\nabla\mathbf u)^T].
$$
The local momentum balance is the <Navier-Stokes equation>:
$$
\boxed{\rho\left(\frac{\partial\mathbf u}{\partial t}+\mathbf u\cdot\nabla\mathbf u\right)=-\nabla p+\eta\nabla^2\mathbf u+\rho\mathbf g.}
$$
The left side is density times <material acceleration>. A steady velocity field can still have convective <acceleration> when a moving parcel changes speed or direction; setting $\partial\mathbf u/\partial t=0$ does not eliminate it.
The <Reynolds number> $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscous effects. At low <Reynolds number>, the fluid's own inertia can often be neglected, giving <Stokes flow>. This does not say that every immersed body has zero <acceleration>: its <mass> and the net pressure, viscous and body <forces> still enter its own <equation of motion>.
For a small rigid sphere moving through a slowly varying viscous background, <Stokes drag> is $-6\pi\eta a(\mathbf v-\mathbf u)$. In still fluid, a simple settling model includes <weight>, <buoyancy> and drag:
$$
m\dot v=(m-\rho_fV_s)g-6\pi\eta av.
$$
Ignoring fluid added mass and unsteady history forces, this gives relaxation time $m/(6\pi\eta a)$ and <terminal velocity>
$$
\boxed{v_t=\frac{2a^2(\rho_s-\rho_f)g}{9\eta}.}
$$
At the terminal <velocity> the net <force> and <acceleration> vanish, while viscous drag and gravity remain individually nonzero. The dissipated <power> $6\pi\eta av_t^2$ is supplied by the loss of gravitational <potential energy> relative to buoyant support. The creeping-flow and quasisteady-drag assumptions must be checked before applying this expression at large speed or immediately after a rapid start.
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