Derive the third equation of motion (v² = u² + 2as) graphically.

PROMPT: A velocity-time graph showing uniform acceleration. The y-axis represents velocity (v) and the x-axis represents time (t). A straight line starts from initial velocity 'u' on the y-axis at t=0 and goes up to final velocity 'v' at time 't'. A horizontal dashed line extends from 'u' to the point (t,u) and a vertical dashed line extends from 'v' to the point (t,v). A point 'A' is at (0,u), 'B' is at (t,v), 'C' is at (t,u), and 'D' is at (0,0). The area under the graph OABD is calculated as the area of a trapezium.

Consider an object moving with uniform acceleration 'a'. Let its initial velocity be 'u' at time t=0, and its final velocity be 'v' at time 't'.
From the velocity-time graph, the displacement (s) is given by the area under the v-t graph, which is the area of trapezium OABD.
Area of trapezium = (1/2) × (Sum of parallel sides) × height
s = (1/2) × (OA + BD) × OD
From the graph, OA = u, BD = v, and OD = t.
So, s = (1/2) × (u + v) × t --- (Equation 1)
From the first equation of motion, we know that v = u + at, which can be rearranged to find t:
t = (v - u) / a --- (Equation 2)
Substitute the value of 't' from Equation 2 into Equation 1:
s = (1/2) × (u + v) × [(v - u) / a]
s = (1/2a) × (v + u)(v - u)
Using the algebraic identity (x + y)(x - y) = x² - y²:
s = (1/2a) × (v² - u²)
2as = v² - u²
Therefore, v² = u² + 2as.