2.4.1 Terminal Velocity

In practice, a falling object experiences other forces besides the gravitational pull mg. This is especially so if the object is very light, or very bulky, or has reached sufficiently high speed such that air resistance R is no longer negligibly small (compared to gravitational pull).

With air resistance, the object will not be a free falling at a constant acceleration of 9.81 m s^-2.

$\begin{aligned}{{F}_{{net}}}&=mg-R\\a&=\frac{{{{F}_{{net}}}}}{m}=\frac{{mg-R}}{m}=g-\frac{R}{m}\text{ }\Rightarrow \text{ }a<g\end{aligned}$

Notice that while the downward weight is fixed, the upward air resistance increases with the velocity. As a result, as the object gains speed, air resistance increases continuously → net force decreases continuously → acceleration decreases continuously. At a sufficiently high speed, air resistance matches weight → net force is zero → acceleration becomes zero! The object has reached its terminal velocity!

The a-t, v-t and s-t graphs (↓ +ve sign convention) of such a motion are as follow:

a-t graph

The a-t graph is a curve that starts at 9.81 m s^-2 and decreases towards zero asymptotically.
As discussed earlier, this is due to increasing air resistance causing the net force to decrease with time.
Note that $a=g$ at $t=0$ . This is because at this instant, the object’s velocity is still zero, meaning there is no air resistance, and the resultant acceleration is g.

v-t graph

the v-t graph is a curve that starts from zero and increases towards the terminal velocity v_t asymptotically.
Since the gradient of a v-t graph corresponds to the acceleration, it is steepest at $t=0$ and becomes flatter and flatter as the acceleration decreases towards zero asymptotically.
Note that the gradient at $t=0$ is equal to 9.81 m s^-2. This is because at the initial point, the object’s velocity is still zero, meaning there is no air resistance, and the resultant acceleration is g.

s-t graph

the s-t graph begins as a quadratic curve but eventually approaches a straight line.
Since the gradient of the s-t graph corresponds to the velocity, its gradient starts off completely horizontal (because $u=0$ ), becomes steeper and steeper as the velocity increases, but eventually stops steepening to become a straight line (because $v={{v}_{t}}$ ).