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how is equation if motion is formed in graphical method
Asked by poornimaram165 | 21 Jun, 2022, 09:16: PM
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 tis 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

Answered by Thiyagarajan K | 21 Jun, 2022, 10:37: PM

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