Physics, asked by aditya316, 11 months ago

derivation of equations of motion​

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Answered by vinitharajnair
3

Answer:

Answer:

 \sf Hlo \: mate \: here's \: your \: answer \: Hope \: it \: helps!

\huge \sf Equations \: of \: Motion

 \sf Consider \: a \: body \: moving \: with \: initial \: velocity \: 'u' \: changes \: it's \: velocity \: to

 \sf 'v' \: after \: 't' \: seconds. \: Let \: 's' \: is \: the \: displacement \: and \: 'a' \: is \: the

 \sf acceleration \: of \: the \: body.

From the defenition of acceleration

a = Change in velocity / time

= (v- u )/ t

by cross multiplication

. at = v - u

 \sf \implies v = u + at

2) displacement (s) = average velocity × time

S = (u + v /2) × t

= [u + (u + at )] / 2 ×t

=( 2u + at ) / 2. × t

= 2ut/2 + 1/2 ×  {at}^{2}

 \sf \implies S \: = \: ut \: + \: \frac{1} {2} {at}^{2}

3). v = u + at

by squaring on both sides

 \sf {v}^{2} \: = \: ({u \: + \: at})^{2}

=  \sf {u}^{2} \: + \: 2uat \: + \: {a}^{2} {t}^{2}

 \sf \implies {v}^{2} \: = \: {u}^{2} \: + \: 2as

Hope it helps you.....

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