Derivation of 3rd Equation of Motion (Graphical Method)

Derivation of 3rd Equation of Motion 

(Graphical Method)

Q.:	Derive first equation of motion?
Ans.	Consider a body moving with initial velocity vi in a straight line with uniform acceleration a. Its velocity becomes vf after time t. The motion of body is described by speed-time graph as shown in the figure below The slope of line AB is acceleration a. The total distance ‘S’ travelled by the body is equal to the total area    OABD under the graph. That is Total distance = S = Total area OABD under the graph Total distance = S = Area of Trapezium OABD \[Total Distance=S=\frac{1}{2}(Sum of Paralell sides)(Distance between Paralell sides)\] \[Total Distance=S=\frac{1}{2}(OA+BD)(OD)\] \[2S=(OA+BD)(OD)\] Multiplying both sides of above equation by  BC/OD \[2S(\frac{BC}{OD})=(OA+BD)(OD)(\frac{BC}{OD})\] \[2S(\frac{BC}{OD})=(OA+BD)(BC)\] \[2S(\frac{BD-CD}{OD})=(OA+BD)(BD-CD)\] ----- (i) From the above graph OA= vi BD= vf CD= vi OD=  t  Putting these values in equation (i) we get \[2S(\frac{v_{f}-v_{i}}{t})=(v_{i}+v_{f})(v_{f}-v_{i})\] \[2S(\frac{v_{f}-v_{i}}{t})=(v_{f}+v_{i})(v_{f}-v_{i})\] \[2S(\frac{v_{f}-v_{i}}{t})=(v_{f}^{2}-v_{i}^{2})\] From 1st equation of motion  vf  - vi  = at Putting this value of  vf  - vi  in the above equation we get \[2S(\frac{at}{t})=(v_{f}^{2}-v_{i}^{2})\] \[2aS=(v_{f}^{2}-v_{i}^{2})\] This is the Third (3rd) equation of motion.