A marble sinking through honey speeds up only briefly, then glides down at a steady pace. The fluid pushes back harder the faster the marble moves. For small, slow objects in thick fluids that drag is close to proportional to the speed, $F_D = -bv$. Newton's second law then becomes a differential equation, and the Separable Equations method solves it. It builds on Motion Under a Time-Varying Force, with a force that depends on the velocity instead of on the time.
Why it works
Take down as positive. $$m\dfrac{dv}{dt} = mg - bv.$$ The speed stops changing when $mg = bv$, at the terminal speed $v_T = mg/b$. Rewrite the equation as $\dfrac{dv}{dt} = \dfrac{b}{m}(v_T - v)$ and separate: $$\int_0^v \dfrac{dv'}{v_T - v'} = \int_0^t \dfrac{b}{m}\,dt' \;\Rightarrow\; -\ln\dfrac{v_T - v}{v_T} = \dfrac{t}{\tau},$$ with the time constant $\tau = m/b$. Solve for $v$.
Key idea
Dropped from rest with drag $-bv$: $$v(t) = v_T\left(1 - e^{-t/\tau}\right), \qquad v_T = \dfrac{mg}{b}, \quad \tau = \dfrac{m}{b}.$$ The gap to terminal speed shrinks by a factor of $e$ every $\tau$: 63% of $v_T$ after one $\tau$, 95% after three, 99% after about 4.6. It never quite gets there. Coasting with no other force: $v = v_0e^{-t/\tau}$.
Worked example: Speed after a while
A $0.1\ \text{kg}$ sphere sinks through oil with $b = 0.5\ \text{N·s/m}$ (ignore buoyancy). Dropped from rest, how fast is it moving after $0.4\ \text{s}$?
- $v_T = \dfrac{(0.1)(9.8)}{0.5} = 1.96\ \text{m/s}$ and $\tau = \dfrac{0.1}{0.5} = 0.2\ \text{s}$.
- $$v(0.4) = 1.96\left(1 - e^{-2}\right) = 1.96(0.8647) = 1.69\ \text{m/s}$$
Answer: $1.69\ \text{m/s}$ (86% of terminal speed).
Worked example: How long to get close
For the same sphere, when does it reach 90% of $v_T$?
- $1 - e^{-t/\tau} = 0.9$ gives $e^{-t/\tau} = 0.1$.
- $t = \tau\ln 10 = 0.2(2.303) = 0.461\ \text{s}$.
Answer: $0.461\ \text{s}$.
Worked example: Coasting to a stop
A $2\ \text{kg}$ puck glides on grease at $5\ \text{m/s}$ with drag $-bv$, $b = 1\ \text{N·s/m}$. How far does it go in total?
- $v(t) = 5e^{-t/2}$ with $\tau = m/b = 2\ \text{s}$.
- $x_{\text{total}} = \int_0^{\infty} 5e^{-t/2}\,dt = 5 \times 2 = 10\ \text{m}$.
- It never quite stops in time, but the distance is finite.
Answer: $10\ \text{m}$.
Common mistake
Two misreadings of terminal speed. First, the object does not stop speeding up because gravity gets weaker; $mg$ is the same all the way down. The drag grows until it matches $mg$, so the net force fades. Second, it does not reach $v_T$ at some finite time and then stop changing; it approaches it exponentially. Check with the formula: $1 - e^{-t/\tau}$ is less than 1 for every finite $t$.
For a small sphere moving slowly through a fluid, Stokes' law gives $b = 6\pi\eta r$, where $\eta$ is the fluid's viscosity and $r$ the radius. That is how the terminal speed of tiny droplets is used to measure their size. Fast objects in air feel Quadratic Drag instead.