Equation 3 is d = v1Δt + ½aΔt²
To help us understand better, let's picture a trapezoid on a graph paper. The equation will be obtained once we find the area of that trapezoid. An important information that we know is that the area of the whole trapezoid will eventually end up to be the total displacement represented on the d-t graph that is correlated with the v-t graph that we are using. In order to obtain the area of the trapezoid, we can calculate the areas of the triangle and of the rectangle separately and later add them. Knowing that the equation used to calculate the area of a triangle is [A = bh/2], we can conclude that on the graph the equation would look like [d = ½(v2 - v1)Δt]. However, to make this equation much simpler, we can replace [v2 - v1] with [aΔt]. Such substitution is able to take place because from the equation used to obtain the slope of a graph [a = (v2 - v1) / Δt], we can further isolate and find [aΔt = v2 - v1]. Similarly, we use our knowledge of the area of a rectangle, which is (A = lw), to calculate the area of the rectangle on the graph, which turns out to be (d = v1Δt). Now that we have the equations to calculate both the areas, we can add them.
Final product - d = area of the rectangle + area of the triangle
= v1Δt + ½(at)Δt
= v1Δt + ½aΔt²
Equation 4 is d = v2Δt - ½aΔt²
Unlike the method used for the derivation of the previous equation, we obtain the area of a large rectangle that is extended from the existing trapezoid; then we must subtract the area of a triangle from that area. Again, we use our knowledge of how to obtain area of certain polygons, in this case that of a triangle and a rectangle. Knowing that the area of a rectangle is equal to the product of the length and the width, we can say that the area of the large rectangle is [d = v2Δt]. The area of a triangle would be [A = bh/2], which on the graph would be equal to [d = ½(at)Δt].
Final product - d = area of the large rectangle - area of the triangle
= v2Δt - ½(at)Δt
= v2Δt - ½aΔt²
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