Equation four can be derived from the velocity time graph by deriving the first two equations and then manipulating equation 2 so that it can be substituted into equation 1. From equation 2, the V1 value is isolated.
from equation 2
a = (V2 - V1) / Δt
a . Δt = V2 - V1
plug this into equation 1
d = 1/2 (V1 + V2) Δt
d = 1/2 (V2 - a . Δt + V2)Δt
d = 1/2 (2V2 - a . Δt) . Δt
d = V2 . Δt - 1/2 . a . Δt²
This equation is only applicable in problems involving constant acceleration, because it was derived from a graph involving constant acceleration.
Equation 4 can be proved from the velocity time graph.
The area from the graph to the x axis is equal to the total distance.
If velocity two is more than velocity one, then by calculating V2 . t we find the area of a rectangle that would give the distance traveled at a constant velocity of the faster speed.
However the speed was not always the faster speed. It started of slow and accelerated from V1 to V2.
Therefore we need to subtract the area of a triangle that is not part of the graph. This area is equivalent to 1/2 of the rectangle formed between the difference of V2 and V1 multiplied by the time taken.