how is equation if motion is formed in graphical method

Let us consider motion is one dimensional with uniform acceleration.

 

In uniform accelrated motion, velocity increases linearly with time.

 

Velocity v at a time t will be

 

v = u + k t

 

where u is intial velocity at t = 0 and k is constant

 

Velocity that varies linearly with respect to time is plotted in above figure.

 

Velocity is u at t = 0  and velocity is v at time t

 

slope = dv/dt = (v-u) / t

 

dv/dt = rate of change of velocity with respect to time = acceleration a

 

Hence we get ,  acceleration a = ( v - u ) / t

 

from above expression we get , v = u + ( a t ) ....................... (1)

 

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In an uniform accelerated motion, let u be initial velocity at time t₁ and v be velocity at later time t₂ .

 

Graph plotted above shows the variation of velocity with respect to time along the line AB .

 

In velocity-time graph , distance travelled from time t₁ to time t₂  is area under the line AB bounded by time axis as shown in figure.

 

Area ABCD under the line AB is a trapezium .

 

Area of trapezium ABCD = (1/2) ( AC + BD ) × CD = (1/2) ( u + v ) (t₁ - t₂ )

 

Hence distance S travelled between time t₁ and time t₂ is given as

 

S = (1/2) ( u + v ) (t₁ - t₂ )  ..................... (2)

 

If we substitute v from eqn.(1) and consider time duration t = ( t₁ - t₂ ) , then we have

 

S = (1/2) [ u + u + (a t ) ] × t

 

Hence , we get ,  S =  ( u t ) + [ (1/2) a t² ] .........................(3)

 

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Let us substitute time duration t = ( t₁ - t₂ ) in eqn.(2) using eqn.(1) as

 

t = ( v - u ) / a

 

Then we have

 

S = (1/2) ( u + v ) [ (v-u) / a ]

 

By simplifying above expression, we get

 

v² = u² + ( 2 a S )  ..................................................(4)

 

Eqn.(1) , (3) and (4) are known as equation of motion in one dimension with uniform acceleration