Question

$\large\sf \underline{question}$
Derivate the 2nd equation of motion, by graphical method.

Note!
Spammed answer(s) will be deleted on the spot ! ​

Answer 1

$\underline{\underline {\sf{\maltese \: \: Derivation \: \: of \: \: 2nd \: \: equation \: \: of \: \: motion }}}:$

Here ,

• $\tt \: initial \: velocity = u$

• $\tt \: final \: velocity = v$

• $\tt \: acceleration = a$

• $\tt \: distance = s$

• $\tt \: time = t$

Now ,

$\tt\implies \: Distance \: covered = Area \: covered \: in \: the \: figure$

$\tt\implies \: Area \: of \: OABC = Area \: of \: OADC + Area \: of \: ABD$

$\tt\implies \: s = ( OA \times OC ) + ( \dfrac { 1 } { 2 } \times AD \times BD )$

$\tt\implies \: s = ( ut ) + ( \dfrac { 1 } { 2 } \times { at }^{ 2 } )$

$\tt\implies \: s = ut + \dfrac { 1 } { 2 } \times { at }^{ 2 }$

Answer 2

$\Large \underline{\tt{\underline{ ✰ \: Tᴏ \: Dᴇʀɪᴠᴀᴛᴇ \: : - }}}$

$\qquad \: \tt{ \underline{❖\:\:Sᴇᴄᴏɴᴅ\: ᴇǫᴜᴀᴛɪᴏɴ\:ᴏғ \: ᴍᴏᴛɪᴏɴ \:ʙʏ\:ɢʀᴀᴘʜɪᴄᴀʟ\:ᴍᴇᴛʜᴏᴅ.}}$

$\Large \underline{\tt{\underline{ ✰ \: Exᴘʟᴀɴᴀᴛɪᴏɴ \: : - }}}$

$\large \underline{ \frak{✠\:Lᴇᴛ, \: : - }} \begin{cases}\sf{\bullet\:(u)\:=\:Iɴɪᴛɪᴀʟ\: Vᴇʟᴏᴄɪᴛʏ\:ᴀᴛ\:(t\:=\:0)\:=\:OA}\\ \sf{\bullet\:(v)\:=\:Fɪɴᴀʟ\: Vᴇʟᴏᴄɪᴛʏ\:ᴀᴛ\:(t\:=\:t)\:=\:BC}\\ \sf{\bullet\:(t)\:=\:Tɪᴍᴇ\:=\:OC}\\ \sf{\bullet\:(a)\:=\:Aᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ}\\ \sf{\bullet\:(s)\:=\:Dɪsᴛᴀɴᴄᴇ\:Cᴏᴠᴇʀᴇᴅ\:ᴀᴛ\:(t\:=\:t)} \end{cases}$

$\tt \underline{ : \implies{Distance \: covered \: (s) \: at \: time \: (t \: = \: t) \implies \: Area \: of \: figure \: \bf (OABC) }} \\\\ \underline{ \tt : \implies{Distance \: covered \: (s) \: at \: time \: (t \: = \: t) \implies \: Area \: of \: figure \: \bf (OADC) \: + \tt \:Area \: of \: figure \: \bf(ABD) }}$

$\sf : \implies \: s \: = (OA \: \times \: OC) \: + \: ( \dfrac{1}{2} \: \times \: AD \: \times \: BD) \\\\ \sf : \implies \: s \: = (u \: \times \: t) \: + \: ( \dfrac{1}{2} \: \times \: t \: \times \: at) \\\\ \bigstar\:\underline{ \boxed{ : \implies { \purple{ \frak{\: s \: = \: ut \: + \: \frac{1}{2} \: at {}^{2} }}}}}\:\bigstar$

$\underline{ \boldsymbol { \therefore \: Hence,} \rm \: the \: second \: equation \: of \: motion, \: is \: derived \: by \: graphical \:method.}$