Drag and Terminal Speed

Weight is constant near the ground and kinetic friction barely changes with speed, so both give constant acceleration. Drag does not. A fluid resists an object more strongly the faster it moves through it, so the resistive force is a function of v. Write the second law along the downward axis and you get m dv/dt = mg − Fdrag(v), which is a differential equation in v alone. From the first instant, every constant-acceleration equation is wrong. The acceleration is largest at release, when the object is slow and drag is small, and it shrinks as the object speeds up.

Linear drag, Fdrag = bv, comes from viscous shear in the layer of fluid dragged along by the surface. For a sphere, Stokes' law fixes the constant: b = 6πηr, with η the dynamic viscosity in Pa s. Quadratic drag, Fdrag = ½ρCdAv², comes from accelerating fluid out of the way and leaving a turbulent wake, so it depends on the fluid density ρ, the frontal area A, and a dimensionless drag coefficient Cd near 0.5 for a smooth sphere and near 1 for a person falling flat. The Reynolds number Re = ρvL/η, with L the diameter for a sphere, picks the regime: below about 1 the flow is viscous and linear drag holds; by a few hundred the wake dominates and quadratic drag holds; between them neither model is clean.

Set the acceleration to zero and solve for v. Linear drag gives bvₜ = mg, so vₜ = mg/b. Quadratic drag gives cvₜ² = mg, so vₜ = √(mg/c). The mass dependence differs: at fixed size, doubling the mass doubles vₜ under linear drag but multiplies it by only √2 under quadratic drag. For a 75 kg skydiver flat to the wind, with ρ = 1.2 kg m⁻³, Cd = 1.0 and A = 0.70 m², c = ½(1.2)(1.0)(0.70) = 0.42 kg m⁻¹, so vₜ = √(735 ÷ 0.42) = 41.8 m s⁻¹. A 1 mm-radius raindrop, mass 4.19 × 10⁻⁶ kg with A = 3.14 × 10⁻⁶ m² and Cd = 0.5, gives c = 9.42 × 10⁻⁷ kg m⁻¹ and vₜ = 6.6 m s⁻¹, which is what raindrops of that size actually do — and its Re ≈ 900 confirms the quadratic choice.

Divide the equation of motion by m and factor out g. Linear drag becomes a = g(1 − v/vₜ); quadratic drag becomes a = g[1 − (v/vₜ)²]. The single expression a = g[1 − (v/vₜ)ⁿ] carries both. Read it without solving anything. At v = 0 the acceleration is g. While v < vₜ the bracket is positive, so the object keeps speeding up and can never cross vₜ. At v = vₜ the acceleration is exactly zero. Throw the object downward faster than vₜ and the bracket turns negative, so it decelerates onto the same asymptote from above. The v–t graph therefore rises with a steadily flattening slope, or falls onto the same horizontal line, depending on where it starts.

Solving the two equations for release from rest gives v = vₜ(1 − e^(−t/τ)) for linear drag and v = vₜ tanh(t/τ) for quadratic drag, with τ = vₜ/g in both. That τ has a physical reading: the initial slope of the v–t graph is g, so τ is the time the object would need to reach vₜ if drag never came on. The real curve bends away from that tangent immediately. Linear drag reaches 63% of vₜ at t = τ and 95% at 3τ. Quadratic drag is faster, at 76% at τ and 96% at 2τ. For the skydiver, τ = 41.8 ÷ 9.8 = 4.27 s, so the speed is 31.9 m s⁻¹ after one τ and 40.3 m s⁻¹ after 8.5 s. For the raindrop, τ = 0.67 s — it is at terminal speed almost as soon as it forms.

Differentiate a measured v–t trace to get a, then plot a against v and against v². Linear drag gives a straight line on the a–v plot; quadratic drag gives a straight line on the a–v² plot. Both lines must extrapolate to a = g at v = 0, and the horizontal intercept is vₜ on the first plot but vₜ² on the second. Judge by the residuals, not by eye. Using the wrong model is not a small error: Stokes' law applied to that 1 mm raindrop, with b = 6πηr = 3.39 × 10⁻⁷ kg s⁻¹, predicts vₜ = 121 m s⁻¹ instead of 6.6. Both forms also assume a fixed shape and a constant coefficient. Cd drifts with Re and drops sharply near Re ≈ 3 × 10⁵ for a smooth sphere, ρ falls with altitude, and buoyancy is ignored — fair for a body in air, where ρair/ρbody ≈ 0.001, but not for a droplet in water.