How to derive the equation d = v2(delta)t - 1/2a(delta)t^2

Similar to the other equation, d = v1(delta)t + 1/2a(delta)t^2, the equation can be found from a velocity - time graph. First, you have the same graph with the same points. This time, you will find the same area except that you first find the area of a rectangle formed with v2 and (delta)t subtracted by the triangle above the trapezoid. The area of the rectangle is

Area1 = bh

Area1 = (v2)(delta)t

Now, the area of the triangle can be found with the equation Area2 = bh/2. The base of this equation is (delta)t like previously but the height is the difference between v2 and v1, which becomes v2 - v1. Therefore the equation becomes:

Area2 = 1/2(v2 - v1)(delta)t

Now, in order to find the area of the trapezoid, you must subtract the area of the triangle from the rectangle.Therefore, you get

Area = Area1 - Area 2

Area = v2(delta)t - 1/2(v2 - v1)(delta)t

Now label this equation ♦

Previously in class, we learned that acceleration can be derived from the slope of the velocity graph:

a (acceleration) = (v2-v1)/(delta)t

a(delta)t = (v2 - v1)

Now label this ♣

Now substitute ♣ into ♦

d(Area) = v2(delta)t - 1/2 (a(delta)t)(delta)t

d = v2(delta)t - 1/2a(delta)t^2

Thus you derived the equation d = v2(delta)t - 1/2a(delta)t^2.