Using graphical methods derive the equation v2
= u2
+ 2as, where the symbols have their
usual meanings.
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The total distance travelled, S = Area of trapezium OABC.
So, S= 1/2(SumofParallelSides)×Height
S=(OA+CB)×OC
Since, OA = u, CB = v, and OC = t
The above equation becomes
S= 1/2(u+v)×t
Now, since t = (v – u)/ a
The above equation can be written as: see
S= 1/2(u+v)×(v-u)/a
Rearranging the equation, we get
S= 1/2(v+u)×(v-u)/a
S = (v2-u2)/2a
Third equation of motion is obtained by solving the above equation:
v2 = u2+2aS
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