Deriving Equation 3 from the Graph

Equation 3 is: d=vΔt+½aΔt²

From the graph, we have V2, V1, t1, and t2.

In order to calculate the displacement, we have to find the area. We can do this in two ways: we can find the area of the triangle and the square, or we can find the area of the entire trapezoid. For this instance, I am going to find the area of the triangle and the rectangle.

The equation for calculating the area of a rectangle is a times b, where a is the height and b is the width.

From the graph, we can see that a=V1, and b=t2-t1.

So, A=(ab)
      A=V1(t2-t1)
      A=V1(Δt)

Now, we have to find the area of the triangle. The equation is a times b divided by 2, where a is the height and b is the length of the base. From this equation, we can find that it can also be expressed as 1/2ab.

From the graph, the base is Δt, and the height is V2-V1.

So, A=1/2Δt(V2-V1)

From Equation #1, we know that V2-V1= aΔt.

Therefore, A=1/2Δt(aΔt)
                 A=1/2aΔt²

So, because d(displacement)=area
Therefore finally, d=vΔt+½aΔt².