Question

Drive second equation of motion by graphical method

Answer 1

Answer:

Let us consider a body which is moving with the initial velocity vi.

Then the final velocity of the body is vf with the time t in sec.

The distance traveled =S, then the acceleration of the body is indicated as a.

That is,

Initial velocity =vi

Final velocity = vf

Time taken =t

Acceleration=a

Distance=S

So now, let us consider the speed of the body by OA which as initial velocity as vi and OC the speed of the body OC at final velocity vf

Then the slope of the graph AB will represent acceleration of the body with a.

S=area of OABC

S= area of rectangle OADC+ area of triangle ABD

Then

S=(vi)(t)+1/2(AD)(BD)

By substitution we get the above expression to be as,

\bold{S=(v i)(t)+1 / 2 a t^{2}}

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Answer 2

$\huge\mathrm\purple{♡HeYa MaTe♡}$

▶️ We know

distance covered = ( average velocity ) × ( time)

or s = u + v / 2 × t

But v = u + at

substituting the value of V in the equation above , we have

s = u + ( u + at ) / 2 × t

or s = 2u + at / 2 × t = ( u + at / 2 ) t

or s = ut + 1/2 at ^2 $\huge\mathrm\blue{♡Answer♡}$

$\huge\mathrm\pink{♡Hope it helps you♡}$

$\huge\mathrm\orange{♡Mark It as brainliest♡}$