Derive the second equation of motion (s = ut + (1/2)at²) 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 represents displacement. This area is divided into a rectangle OACD and a triangle ABC.

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.
Area of trapezium OABD = Area of rectangle OACD + Area of triangle ABC
Area of rectangle OACD = length × breadth = OA × OD = u × t
Area of triangle ABC = (1/2) × base × height = (1/2) × AC × BC
From the graph, AC = OD = t.
And BC = BD - CD = v - u.
We know from the first equation of motion that v - u = at.
So, Area of triangle ABC = (1/2) × t × (at) = (1/2)at².
Therefore, Total Displacement (s) = ut + (1/2)at².