In the case of free fall, this becomes:
$$\dot{v} = g$$
More Analytic Solutions:
If $F = -bv$, then we can solve for velocity and obtain:
$$v(t)= v_0 e^{\frac{t}{\tau}}$$
But, what about when
$$\int dv = \int \left(\textrm{Something Nasty} \right) dt$$
Numerical Methods
Re-writing our original definition of acceleration, we can say that:
$$\frac{\Delta v}{\Delta t} = a$$
Let's say we know a value for $v_i$
$$v_f - v_i = a \Delta t$$
Then we can just solve this to get a value for $v_f$ at some $\Delta t$.
A plot of a vs. t
The Area under the curve
Numerical Integration
Let's start with $v_i = 0$. Then we can see that
$$v_f = a \Delta t$$
That the area of the first rectangle in the previous graph.
Now we can use that $v_f$ in the original equation as the $v_i$, and we will obtain:
$$v_f = v_i + a \Delta t = a \Delta t + a \Delta t$$
... and so on.
