falling body with air resistance
Drop a feather and a coin in air and the coin wins easily; drop them in a vacuum and they fall together. The difference is air resistance — drag — and it is why a skydiver does not accelerate forever but settles into a steady plunge. Modeling a falling body with air resistance is the classic first-order Newton's-law problem, and it explains terminal velocity beautifully.
Newton's second law says mass times acceleration equals the net force: m dv/dt = mg − (drag). Gravity pulls down with force mg; drag pushes up and grows with speed. The simplest honest choice is drag proportional to velocity, giving m dv/dt = mg − kv, a first-order linear equation. As the body speeds up, the drag kv climbs until it balances gravity, the net force hits zero, and acceleration stops; the velocity then levels off. Solving gives v(t) = (mg/k)(1 − e^(-(k/m)t)) for a body dropped from rest — the speed rises and flattens out toward a ceiling.
That ceiling is the terminal velocity, mg/k, where the forces cancel. The model's honesty hinges on the drag law: drag proportional to v is reasonable for slow, small, or viscous motion, but for a fast skydiver drag scales more like v^2, giving m dv/dt = mg − kv^2 — still first-order, still with a terminal velocity (now sqrt(mg/k)), but nonlinear. Either way, the lesson is the same: speed-dependent resistance turns 'falling forever faster' into 'settling to a steady speed.'
A 70 kg skydiver with linear drag m dv/dt = mg − kv and k = 14 kg/s has terminal velocity mg/k = 70·9.8/14 = 49 m/s. The speed climbs as v(t) = 49(1 − e^(-0.2 t)), reaching about 49 m/s within roughly 20 seconds.
Drag rises with speed until it cancels gravity; speed then levels off.
Drag proportional to v is a simplification; for fast, bluff bodies in air the realistic law is closer to drag proportional to v^2, which is nonlinear but still has a terminal velocity. Pick the law to match the regime.