Formula 1:
This formula comes directly from the graph.
To find acceleration on a v -> t graph, you find the slope.
The slope can be given by the two coordinates on the graph.
a = rise/run
= ∆v / ∆t
= (v2 - v1) / ∆t
therefore: a∆t = v2 - v1
Formula 2:
To find d on a v-> t graph, you find the area according to the x-axis.
Once again, the coordinates on the graph can help you find the area.
The area of a trapezoid is: A = (a+b)h /2
d = (v1 + v2)∆t / 2
therefore: d = 1/2(v1 + v2)∆t
Formula 3:
Formula 3 & 4 must have a and d, because they are already proven.
From Formula (1), isolate v2.
v2 = a∆t + v1 ... let it be represented by :)
Sub :) into Formula (2).
d = 1/2(v1 + :) )∆t
d = 1/2(v1 + a∆t + v1)∆t
d = 1/2∆t(2v1 + 2∆t)
d = v1∆t + 1/2a∆t²
therefore: d = v1∆t + 1/2 a∆t²
Formula 4:
Formula 4 is very similar to 3.
From Formula (1), isolate v1 instead of v2 now.
v1 = a∆t - v2 ... let it be represented by :(
d = 1/2(:( + v2)∆t
d = 1/2(-a∆t + v2 + v2)∆t
d = 1/2∆t(-a∆t +2v2)
d = v2∆t - 1/2a∆t²
therefore: d = v2∆t - 1/2a∆t²
Formula 5:
Formula 5 comes from Formula 1. And there are many ways to find it.
Method 1)
Isolate ∆t from Formula 1.
∆t = v2 - v1/a
d = 1/2(v1 + v2)(v2 - v1/a)
ad = 1/2(v2 + v1)(v2 - v1)
2ad = v2² - v1²
therefore: v2² = v1² + 2ad
Method 2)
Isolate ∆t from Formula 1.
∆t = v2 - v1/a ... we'll call this A
Isolate ∆t from Formula 2.
∆t = 2d/v1 + v2 ... we'll call this B
A = B because ∆t = ∆t
v2 - v1/a = 2d/v1 + v2
(v2 - v1)(v1 + v2) = 2d(a)
v2² - v1² = 2ad
therefore: v2² = v1² + 2ad
