Deriving equations from the graph but flipped because I got my trapezoid backward, yah :)

When given the velocity graph, we already know that by calculating the area of the shape in respect of the x axis, we will get the displacement.

Generally, the shapes are rectangles, triangles and the combination of both - trapezoid.

We know that the formula for calculating the area for rectangle is:

A = LW,

and formula for triangle is:

A = 1/2 BH

and trapezoid:

A=(a+b)H/2

But sometimes, the trapezoid can also be seen as a triangle and rectangle stacked on top of each other. Therefore, another way of calculating the area of the trapezoid can be: Area of triangle + Area of Rectangle

A = LW + 1/2BH

In this sketch I drew on paint (:, the graph is a typical velocity time graph. As ou can see, the trapezoid is divided into 2 parts - a triangle and a rectangle.

So first, the base of triangle as we can see, is time(t), the height is the difference between v1 and v2. So the area of the triangle is:

A = 1/2(t)(v1 - v2)

For the area of the rectangle, obviously the length is t, and the width is v2, so the area is:

A = v2 * t

By adding the sum of these areas, we get the area of the trapezoid, or otherwise as we know the displacement:

D = 1/2(t)(v1 - v2)+v2*t (2)

Now from previous knowledge, we already have equation 1):

v2=v1+at (1) => v1=v2-at (3)

Sub (3) into (2):

D=1/2t(v2-at-v2)+v2*t

D=(v2)(t)+1/2at² 

And Voila! Equation 3 from the Big Five!

To derive equation 4, the idea is about the same, except this time, you look at the graph a little differently

You treat the trapezoid as a bigger rectangle, but without the shaped triangle. So another way to calculate its area is by calculating the area of the rectangle, and subtract the area of the imaginary triangle from it:

A = LW - 1/2(BH)

First, the length of the rectangle as we can see is t, the width is v1:

Area of Rectangle: v1 * t

Secondly, the base of the triangle is again t, the base as you can see by observing more closely, is (v1-v2).

Area of Triangle: 1/2(t) * (v1-v2)

Therefore, area of trapezoid: v1 * t - 1/2(t) * (v1-v2) (1)

Once again we apply the first equation that we already know of:

v2 = v1 + at (2)

Sub (2) into (1)

D = v1 * t - 1/2(t) * (v1 - v1 - at)

D = (v1)(t) - 1/2at²

And Ta da! Equation 4!

(Please note that v2 and v1 in the equations are opposite of the ones in the textbook, this is because the longer and shorter side of the trapezoid I have is directly opposite of the ones we did in class)