Derive the second equation of motion by graphical method


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Derive the second equation of motion by graphical method​

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To Derive 2nd equation of motion [Graphically]

★ s = ut + ½ at²

Consider the velocity time graph of a body shown in attachment. The body has an initial Velocity u at point A and then velocity changes from A to B at a uniform rate in time t. There's uniform acceleration a from A to B,and after time t the final Velocity is v which is equal to BC. The time t is represented by OC. Through construction we drew perpendicular CB from point C,and AD parallel to OC. BE is perpendicular to the point B to OE.

Here,

• Initial Velocity (u) = OA

• Final Velocity (v) = BC

• Time (t) = OC

Distance Covered = Area of the figure OABC

Area of OABC = Area of OADC + Area of ABD

→ Distance (s) = Area of OADC + Area of ABD

→ s = (OA × OC) + (½ × AD × BD)

→ s = (u × t) + (½ × t × at)

s = ut + ½ at²

\therefore The 2nd equation of motion is derived here graphically.


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