Physics, asked by vamritaeunameun, 1 year ago

PLEASE PROOF ALL EQUATIONS OF MOTION.
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Answers

Answered by Anonymous
9

\rm\color{palevioletred}{\underline{1.\: First \:equation \:of \:motion \: :- }}

v = u + at

 \sf{ \: we \: know \: acceleration(a) \:  =  \frac{change \: in \: velocity}{time \: taken} }

 \sf{ a \:  =  \frac{final \:velocity \: - \: initial \:velocity}{time \: taken} }

 \sf{or \: \:  \:  \:  a =  \frac{v - u}{t} } \\  \sf{or \:  \: at = v - u} \\   \fbox{\sf{ \therefore   \: v = u + at}} \: ....(1)

\rm\color{palevioletred}{\underline{2.\: Second \:equation \:of \:motion \: :- }}

 \sf{s =ut + \frac{ 1}{2}{at}^{2} }

We know ,

\sf{\implies Distance \:travelled = Average\: Velocity × Time}

</p><p>\sf{ Distance  \: travelled =\frac{Initial \: Velocity+Final \: Velocity}{2}×Time}</p><p>

\sf{s=\frac{v+u}{2}×t}

From equation (1) \impliesv = u+at

\sf{\therefore s =\frac{ u+(u+at)}{2}×t}

 \sf{or \:  \: s =\frac{ 2u+at}{2}×t}

 \fbox{ \sf{s =ut + \frac{ 1}{2}{at}^{2} }} \: .....(2)

\rm\color{palevioletred}{\underline{3.\: Third \:equation \:of \:motion \: :- }}

\sf{Distance\: travelled = Average\: Velocity ×\: Time}

\sf{Distance \:travelled\: =\frac{Initial\:Velocity\:+\:Final\:Velocity}{2}×Time}

\sf{ \implies s= \frac{v+u}{2}×t}

From equation (1) \impliesv=u+at

\sf{or \: t= \frac{v-u}{a}}

s= \frac{v+u}{2} × \frac{v-u}{a}

 \implies\sf{s =  \frac{ {v}^{2} -  {u}^{2}  }{2a} }

or v²-u² = 2as

\sf\fbox{{v}^{2}= {u}^{2}+2as }.....(3)


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Answered by Anonymous
2

Explanation:

Refer to the attachment

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