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Derive equations of motion by graphical method.

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An object is in motion with initial velocity u attains a final velocity v in time t due to acceleration a, with displacement S. 

Let us try to derive these equations by graphical method. 

Equations of motion from velocity – time graph:

Graph shows the change in velocity with time for an uniformly accelerated object. The object starts from the point D in the graph with velocity u. Its velocity keeps increasing and after time t it reaches the point B on the graph. 

The initial velocity of the object = u = OD = EA 

The final velocity of the object = v = OC = EB

Time = t = OE = DA 

Also from the graph we know that, AB = DC 

For First equation of motion 

By definition, acceleration = change in velocity / time 

= (final velocity – initial velocity)/time 

= (OC – OD) / OE 

= DC / OE 

a = DC/t 

DC = AB = at

From the graph EB = EA + AB 

v = u + at ….(1) 

This is first equation of motion.

For Second equation of motion 

From the graph the distance covered by the object during time t is given by the area of quadrangle DOEB

s = area of the quadrangle DOEB 

s = area of the rectangle DOEA + area of the triangle 

DAB = (AE × OE) + (1/2 × AB × DA) 

s = ut + 1/2at2 ….(2) 

This is second equation of motion

For Third equation of motion 

From the graph the distance covered by the object during time t is given by the area of the quadrangle DOEB. Here DOEB is a trapezium. 

Then s = area of trapezium DOEB

 = 1/2 × sum of length of parallel side × distance between parallel sides 

= 1/2 × (OD + BE) × OE s 

= 1/2 × (u + v) × t 

Since a = (v – u) / t or t = (v – u)/a 

Therefore = 1/2 × (v + u) × (v – u)/a . 

2as = v = u + 2as v = u + 2as ………(3) 

This is third equation of motion.

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