derive the 2nd equation of motion. graphically and normally.
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Please give the complete question please
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the derivation is
t = 0, initial velocity = u = OA
At t = t, final velocity = v = OC
The distance S travelled in time t = area of the trapezium OABD
s = (1/2) x (OA + DB) × OD
s = (1/2) x (u + v) × t
Since v = u + at,
s = (1/2) x (u + u + at) × t
s = ut + (1/2) at2
PLS MRK AS BRAINLIEST
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