How to find area of intersection of two moving polygons? We have two non-convex polygons:
M: $(x^1, y^1), (x^2, y^2),..., (x^m, y^m)$;
N: $(x^1, y^1), (x^2, y^2),..., (x^n, y^n)$;
They move at speeds:
$v^M=(v^M_x, v^M_y)$;
$v^N=(v^N_x, v^N_y)$;  
How to find moment of time t where area of intersection is maximum?
 A: You can find the time of maximum area of intersection by setting the first polygon's motion vector to zero and the second polygon's motion vector to
$v^N - v^M.$
So now we have reduced the problem to a case where only one polygon is moving. Transform all coordinates so that the motion is in the direction of the positive $x$ axis.
Take the set of all vertices of both polygons.
Through each vertex, construct a line parallel to the $x$ axis.
Use these lines to partition each polygon into trapezoids.
(Some of the trapezoids may be triangles, that is, degenerate trapezoids in which one base has zero length. Since the polygons were initially not convex,
there may be multiple trapezoids from the same polygon between two adjacent lines.)
Now consider the trapezoids in which the second polygon is partitioned,
all moving at the same speed parallel to their bases.
If we consider just one of these trapezoids along with one of the trapezoids of the first polygon,
the area of intersection of the two trapezoids is a piecewise quadratic function of time,
and the pieces of that function begin and end at times when a vertex of one trapezoid coincides with a vertex of the other trapezoid, which times can easily be calculated.
There are four of these times, and depending on the sequence in which they occur (which pairs of coordinates coincided before which other pairs of coordinates) there are just a few cases for deciding which quadratic function represents the area between those times.
Now take the sum of the functions for all pairs of trapezoids (one from the first polygon, one from the second) that can have non-zero area of intersection (that is, their bases are on the same pair of parallel lines).
The result is a single piecewise quadratic function.
Maximize that function.
A: As said in the very good solution given by @David K, one can consider relative motions, i.e. one can assume WLOG that polygon $M$ is fixed, and polygon $N$ moves along the $x$ axis.
What I advise is the following:
1) Solve the particular case of a pair of moving triangles with coordinates :
$(x_k \ ; \ y_k)$ for triangle $M$  and $(x'_k+ vt \ ; \ y'_k)$ for triangle $N$ for a certain horizontal translation  $x=vt$.
The important result is that this function is always continuous and that its graphical reprensentation is made by "gluing together" parabolic arcs (or exceptionaly line segments), like a quadratic spline but without differentiability everywhere.
2)  Now, for the general case, let us triangularize the interiors of polygons  $M$ and $N$ as $\cup_k\mathfrak{T}_k$ and $\cup_{\ell}\mathfrak{T'}_{\ell}$
resp. (this is always possible for any kind of polygon). It is sufficient now, by linearity, to apply the method developed for the particular case of two triangles to the sum of areas of intersections of every triangle of type $\mathfrak{T}_k$ with every triangle of type $vt+\mathfrak{T'}_{\ell}$:
$$f(t)=\sum_{k} \sum_{\ell} area\left(\mathfrak{T}_{k} \cap (vt+\mathfrak{T'}_{\ell})\right)$$ 
As the set of continuous function with compact support and at most second degree polynomials on intervals, $f$ has the same type.
Appendix: Consider the particular case where 


*

*the vertices of $M$ are $A(a,1)$, $B(a+1,0)$, $C(a+1,1)$ (black triangle in Fig. 1)

*the vertices of $N$ are $D(0,1)$, $E(0,0)$, F$(1,0)$ (blue triangle in Fig. 1)
Then, it is not difficult to find that the intersection of triangles $M$ and $N$  (which is a parallelogram) when $N$ slides over $M$ at unit speed has an area given by the quadratic formula $f(t)=t(1-t)$ for $0<t<1$ and $f(t)=0$ elsewhere.
A second example is given below with the superposition of triangle $ABC$ and triangle $FGD$, and a third one by adding the two first examples. Anyhow, we will always obtain a $C^0$ curve with compact support made by piecing together parabolic (or linear) arcs.
Working with examples does not mean that we want to bypass a rigorous proof, but it can be done in a second step. My point here is that ideas are the most important things to convey, and ideas are often understood on examples.

Fig. 1 : Triangle $ABC$ is moving with speed vector $(a,0)$. Square $DEFG$, fixed, is decomposed into 2 triangles. 

Fig. 2 : Sweeping of triangle $ABC$ on square $DEFG$ (see Fig. 1) can be decomposed into sweeping of triangle $ABC$ on triangle $DEF$ (common area represented by the blue curve) and sweeping of triangle $ABC$ on triangle $FGD$ (red curve ; note that the maximum of this curve corresponds to the exact superposition of black and red triangles). Adding contributions yields the black curve.
