Recipe

Find terminal velocity and predict how a falling object approaches it

Use when: you have an autonomous velocity equation $\dfrac{dv}{dt}=g-\dfrac{k}{m}v$ (or any equation of the form $v'=(\text{constant})-(\text{positive constant})v$ ) and an initial speed $v_0$

Goal: the terminal velocity $v^\star$ , confirmation that it's stable, and which direction a given $v_0$ approaches it from

Taught in Terminal velocity with air resistance

Quick recipe

Set the rate to zero and solve for the equilibrium
Confirm the equilibrium is stable
Compare the initial speed $v_0$ to $v^\star$

1
Solve $g-\dfrac{k}{m}v=0$ for $v$ ; call the result $v^\star$ .

Example

For $\dfrac{dv}{dt}=9.8-0.25v$ : $9.8-0.25v=0 \Longrightarrow v=9.8/0.25=39.2$ .

Why: an equilibrium is, by definition, the constant solution where the rate of change is zero — no integration needed, since the equation is already solved for $v'$ explicitly.
2
Check the sign of $\dfrac{d}{dv}\Big[g-\dfrac{k}{m}v\Big]=-\dfrac{k}{m}$ ; since $k,m>0$ this is always negative, so $v^\star$ is always stable for this family.

Example

$k=0.25$ , $m=1$ : the derivative is $-0.25<0$ , confirming stability.

Why: a negative derivative at the equilibrium means arrows point in from both sides on the phase line — exactly the stable case from earlier in the course.
3
Check whether the object starts out slower or faster than the terminal velocity found in Step 1 — that comparison alone determines which direction $v(t)$ moves.

Example

With $v^\star=39.2$ : $v_0=50>39.2$ , so speed decreases toward $39.2$ from above.
- $v_0<v^\star$ — the object is still speeding up; $v(t)$ increases toward $v^\star$ from below.
- $v_0>v^\star$ — the object was launched faster than terminal velocity; $v(t)$ decreases toward $v^\star$ from above.
Why: because $v^\star$ is stable (Step 2), every trajectory moves toward it monotonically from whichever side it started on — it never overshoots.

Common mistakes

Solving $g-\frac{k}{m}v=0$ incorrectly for $v$ , such as reporting $v^\star=g\cdot\frac{k}{m}$ instead of $v^\star=g/(k/m)$ .
Assuming a falling object always speeds up toward terminal velocity — an object thrown downward faster than $v^\star$ actually slows down toward it instead.
Confusing a larger drag coefficient $k$ 's effect on the VALUE of $v^\star$ (smaller) with its effect on stability (the equilibrium is stable either way, just approached faster or slower).
Forgetting that this whole approach assumes drag proportional to speed — a different drag law, such as drag proportional to $v^2$ , gives a different, nonlinear equation with a different-looking equilibrium condition.