Derivaton of 3rd equation of motion graphically
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Derive v2 = u2 + 2as by Graphical MethodVelocity–Time graph to derive the equations of motion.We have just seen that the distance travelled s by a body in time t is given by the area of the figure OABCwhich is a trapezium. In other words,Distance travelled, s=Area of trapezium OABCNow, OA + CB = u + v and OC = t. Putting these values in the above relation, we get: ...... (7)We now want to eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion. Thus, v = u + at (First equation of motion)
And, at = v – u or Now, putting this value of t in equation (7) above, we get: or 2as=v2 – u2 [because (v + u) × (v – u) = v2 – u2]or v2=u2 + 2asThis is the third equation of motion.
And, at = v – u or Now, putting this value of t in equation (7) above, we get: or 2as=v2 – u2 [because (v + u) × (v – u) = v2 – u2]or v2=u2 + 2asThis is the third equation of motion.
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