Physics, asked by bhagyashripotdpc85rh, 1 year ago

derive the three equations of motions by the using graphical method

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Answered by Anonymous
14

Heya !!

The answer goes here...

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Let the angle be x

1st \: equation - v = u + at

Since we know that the slope or gradient of v-t graph shows acceleration. Therefore,

a = tan \: x \\ a = \frac{BC}{AC} \\ a = \frac{BD - OA}{OD} \\ a = \frac{v - u}{t} \\ v = u + at \\ \\ Proved

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2nd \: equation - \: s = ut + \frac{1}{2} {at}^{2}

Since, we know that the area between v-t graph and time-axis represent the displacement of the body in a given interval of time. Therefore,

s = area(ABDOA) \\ s = ar(ABC) + ar(ACDA) \\ s = OA \times OD + \frac{1}{2} \times BC \times AC \\ s = OA \times OD + \frac{1}{2} \times OD \times BC \: \: \: \: \: - - - (i)

In ∆ ABC,

a = tan \: x \\ a = \frac{BC}{OD} \\ BC = a \times od \: \: \: \: \: - - - (ii)

from \: (i) \: and \: (ii) \\ s = OA \times OD + \frac{1}{2} \times OD \times OD \times a \\ s = ut + \frac{1}{2} {at}^{2} \\ \\ Proved

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3rd \: equation - \: {v}^{2} = {u}^{2} + 2as

Since, the area between v-t graph and time-axis in the given time interval represent the displacement. Therefore,

s = area(OABD) \\ s = \frac{1}{2}(OA \times BD) \times OD \: \: \: \: \: - - - (iii)

In ∆ ABC,

tan \: x = \frac{BC}{AC} \\ a = \frac{BD - CD}{OD} \\ OD = \frac{BD - OA}{a} \: \: \: \: \: - - - (iv)

from \: (iii) \: and \: (iv) \\ s = \frac{1}{2} (OA + BD) \times ( \frac{BD - OA}{a} \\ 2as = {BF}^{2} - {OA}^{2} \\ 2as = {v}^{2} - {u}^{2} \\ {v}^{2} = {u}^{2} + 2as \\ \\ Proved

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Hope it helps..

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Anonymous: tnx :)
Answered by Anonymous
8
Hope it helps you !!

Dude sorry these are from my class work....
If you think it is cheating then please repory Answer ( Due to shortage of time )
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