How to derive equation 3 and 4 by using a graph? It is simple! imagine a trapezoid on a graph. Since the trapezoid is composed of a right triangle and a rectangle shaped figure, we can actually simply add the area of the right triangle and the rectangle for equation three:

Area for rectangle:  (v1)(t)

Area for right triangle:   (v2-v1)(t)/2

So, after combining these two results, the equation would be: 

d = v1t + ½(v2-v1)t

Because at = v2-v1 the next step would be substituting this equation into the one above. So it becomes:  d = v1t + ½at(t). As we continue, the final equation which is the third equation would be d = v1t + ½at²


Equation Four


This time, we change our steps for finding the area of the trapezoid. We previously used addition to find the area, therefore this time we will be using subtraction. We first calculate the total area of a rectangle which is the trapezoid, but we see it as a rectangle. The equation will be: (v2)(t). In order for us to get the correct area of the trapezoid, we have to subtract the excessive area we included. So the excessive area would be: (v2-v1)(t)/2

Again, since at = v2-v1, we substitute this into the result we got from above and the equation would be like: d = v2t-½(v2-v1)t, then, 

d = v2t-½at(t), continue to d = v2t-½at², and we are done!