terminal velocity
Why does a falling raindrop, or a skydiver before the parachute opens, eventually stop speeding up and fall at a steady rate? Because air resistance grows with speed, and at some speed it grows just enough to exactly cancel gravity. With no net force left, there is no acceleration, and the object cruises at that constant speed. That speed is the terminal velocity.
It is the equilibrium of the falling-body equation. Take m dv/dt = mg − kv: set the acceleration dv/dt to zero, which means the right side is zero, so mg − kv = 0 and v = mg/k. At this velocity the forces balance for good; if the object were somehow moving faster, drag would exceed gravity and slow it back down, so the terminal velocity is a stable equilibrium that all falling motions approach. With the more realistic quadratic drag mg − kv^2 = 0, the terminal velocity is sqrt(mg/k).
Terminal velocity is the cleanest everyday example of an equilibrium solution and of the qualitative idea that solutions settle toward a stable steady state. It explains why heavy and light objects reach different cruising speeds, why a flat sheet of paper drifts slowly while a crumpled ball drops fast, and why parachutes work — opening one dramatically increases k, slashing the terminal velocity to a survivable landing speed.
A skydiver with mg = 686 N and linear drag constant k = 14 kg/s has terminal velocity mg/k = 49 m/s. Open the parachute and k might jump to 140 kg/s, dropping the terminal velocity tenfold to about 4.9 m/s — a gentle landing.
Set acceleration to zero: terminal velocity is the equilibrium where drag cancels gravity.
Terminal velocity is approached but, like any exponential settling, never reached in finite time — the object gets arbitrarily close to mg/k without ever exactly equalling it.