How to derive the equation d = v1(delta)t + 1/2a(delta)t^2

Velocity Graph with points (v1, t1) and (v2, t2)

First, let's label the first point (v1, t1) where v1 is the velocity at a certain point and t1 is the time at which the velocity v1occurs. Second, let's label the second point (v2, t2) where v2 is another velocity at a certain point and t2 is the time at which v2 occurs. In order to translate the graph into a distance-time graph by finding the total displacement, the area formed with the x-axis must be found. Divide the trapezoid formed into 2 sections, the rectangle and the triangle. The area of the rectangle can be found with the equation Area1 = bh where b is the base, the change in time between t2 and t1 named (delta)t, and where h is the height, v1. Therefore the

Area1 = v1(delta)t

The area of the triangle can be found with the equation Area2 = (bh)/2, where the base is (delta)t and the height is v2 - v1 since you are trying to find the height between v2 and v1. Therefore, your equation is

Velocity Graph with areas indicated of the (tri + rect)angles

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

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

Now, let's combine the 2 areas and label it ☺: d(Area) = v1(delta)t + 1/2(v2-v1)(delta)t

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

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

Now label this ☻                                           a(delta)t = (v2 - v1)

Now substitute ☻ into ☺:                d = v1(delta)t + 1/2(a(delta)t)(delta)t

d = v1(delta)t + 1/2(a (delta)t^2)

d = v1(delta)t + 1/2a(delta)t^2

Thus you derived the equation d = v1(delta)t + 1/2a(delta)t^2.