Derivation of 2nd Equation of Motion (Graphical Method)

 Derivation of 2nd Equation of Motion 

(Graphical Method)




Q.:

Derive Second (2nd) 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 rectangle OACD + Area of triangle ABC   ----------------- (1)

From the graph above

Area of rectangle OACD = OA x OD --------------------  (i)

From the above graph

OA=   vi    and       OD= t

Putting these values in equation (i), we get

Area of rectangle OACD =  vi x t

From the graph above 

\[Area of triangle=\frac{1}{2}(\overline{AC}(\overline{BC})\] -- (ii)

From the graph above

AC= t  and BC= vf  - vi

Putting these values in equation (ii), we ger

\[Area of triangle=\frac{1}{2}(t)(v_{f}-v_{i})\]

From 1st equation of motion

 vf  - vi  = at

Putting this value of  vf  - vi  in the above equation we get

\[Area of triangle=\frac{1}{2}(t)(at))\]

\[Area of triangle=\frac{1}{2}at^{2}\]

Putting values of Area of rectangle OACD  and Area of triangle ABC in equation (1) 

we will get

\[S= v_{i}t+\frac{1}{2}at^{2}\]

This is the second (2nd) equation of motion.




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