A resistive force opposes motion relative to a surrounding medium. In a common linear drag model, the force is proportional to velocity:

$$\boxed{\mathbf F_d=-b\mathbf v},$$

where $b>0$ is a drag coefficient. The minus sign makes the force point opposite the velocity.

Linear drag is useful when resistance is approximately proportional to speed over the range being modeled.

Falling under gravity with linear drag

Take downward as the positive direction. For an object of mass $m$ falling with downward speed $v\ge0$, gravity is positive and drag is negative:

$$m\frac{dv}{dt}=mg-bv.$$

A terminal velocity occurs when acceleration becomes zero:

$$0=mg-bv_t.$$

Therefore

$$\boxed{v_t=\frac{mg}{b}}.$$

Terminal velocity is not a new force. It is the steady speed at which the downward weight and upward drag balance, so the net force vanishes.

Exponential approach to terminal velocity

Rewrite the equation as

$$\frac{dv}{dt}+\frac{b}{m}v=g.$$

Define the time constant

$$\boxed{\tau=\frac{m}{b}}.$$

The solution for initial speed $v_0$ is

$$\boxed{v(t)=v_t+(v_0-v_t)e^{-t/\tau}}.$$

If the object is released from rest,

$$\boxed{v(t)=v_t\left(1-e^{-t/\tau}\right)}.$$

Initially, when $v=0$, drag is zero and the acceleration is $g$. As the speed increases, drag grows, the net acceleration decreases, and the velocity approaches $v_t$ asymptotically.

The time constant sets the response scale. After one time constant,

$$v(\tau)=v_t(1-e^{-1})\approx0.632v_t$$

for release from rest.

Worked example

A falling object has

$$m=80,\mathrm{kg}$$

and linear drag coefficient

$$b=40,\mathrm{kg/s}.$$

Its terminal velocity is

$$v_t=\frac{(80)(9.81)}{40}\approx19.6,\mathrm{m/s},$$

and its time constant is

$$\tau=\frac{80}{40}=2.0,\mathrm s.$$

If released from rest, after $4.0,\mathrm s=2\tau$ its speed is

$$v(4)=19.6\left(1-e^{-2}\right)\approx17.0,\mathrm{m/s}.$$

It is already close to terminal speed but has not reached it exactly.

Direction matters

The vector form $\mathbf F_d=-b\mathbf v$ automatically reverses the drag direction when the object reverses its motion. A scalar equation such as $mg-bv$ is valid only after a coordinate direction and velocity sign convention have been chosen consistently.

Linear drag turns Newton's second law into a first-order differential equation and provides a simple example of how a speed-dependent force produces an approach to dynamical equilibrium rather than constant acceleration.