# Confusion with Terminal Velocity

So I know that terminal velocity occurs when velocity becomes constant and the body is no longer accelerating.

However using our normal kinematics equation say a body in terminal velocity, the time it takes for say an object to fall $$10000$$m if the body reaches terminal velocity in between will imply $$a = 0.$$ So $$10000 = \frac{1}{2}gt^2$$. But our acceleration is now $$0$$ so the time is infinite?

Or do we use $$10000 = v_Tt + \frac{1}{2}at^2?$$ Where we now use the terminal velocity as our initial speed $$u$$ in the original equation $$s = ut + \frac {1}{2}at^2?$$ ($$s$$ is displacement).

Where is my understanding wrong here?

The equations you have mentioned are derived for constant acceleration throughout the motion.If the acceleration changes in between, as in your case it suddenly changes to zero you have to apply the equation again for the new set of conditions(new acceleration is zero).But if the acceleration changes continuously with time, acceleration should be expressed as a function of time and integrated twice to get the displacement as a function of time.

The usual way to express the effect of air drag is an acceleration that is some function of velocity, and in the opposite direction to the velocity. So an object falling in air, and choosing "down" as positive, you get this.

$$a = g - D(v)$$

where $$g$$ is gravity, and $$D(v)$$ is the drag as a function of velocity. In some important situations it will simply be some constant times the square of the velocity. (But that is certainly not the only possible situation.)

https://en.wikipedia.org/wiki/Drag_equation

In that case you have the following.

$$a = \frac{dv}{dt} = g - k v^2$$

So if you have $$v=0$$ as your initial condition at $$t=0$$ you can do the integral.

$$\int \frac{dv}{g-kv^2} = t$$

Which will require some partial fractions to complete.

The terminal velocity is such that this is zero. Meaning the following.

$$kv^2 = g$$

Or, solving for $$v$$ you get this.

$$v = \sqrt{\frac{g}{k}}$$

Your kinematics equations only apply to motion without any resistive forces. When there is air resistance, this balances out the gravitational force, and that causes the acceleration to eventually reach zero.

The equation for air resistance is $$kv^2$$ for some constant $$k$$.

Therefore your equation becomes a differential equation. Taking $$y$$ as the height, we can write it as: $$m\ddot y = -mg + k {\dot y}^2.$$ The solution to this is not simply $$y=y_0-gt^2/2$$.