Question Number 200632 by pascal889 last updated on 21/Nov/23
$$\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{derived}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{formular}}\:\boldsymbol{\mathrm{of}} \\ $$$$\:\boldsymbol{\mathrm{motion}} \\ $$
Commented by mr W last updated on 21/Nov/23
$${do}\:{you}\:{really}\:{think}\:{this}\:{is}\: \\ $$$${appropriate}?\:{this}\:{is}\:{like}\:{when}\:{i}\:{ask} \\ $$$${you}\:{to}\:{explain}\:{me}\:{mathematics}. \\ $$$${where}\:{to}\:{begin}? \\ $$
Answered by AST last updated on 21/Nov/23
$${If}\:{you}\:{mean}\:{derive}\:{from}\:“{first}\:{principles}'': \\ $$$$\frac{{dv}}{{dt}}={a}\Rightarrow{dv}={adt};\frac{{ds}}{{dt}}={v}\Rightarrow{ds}={vdt} \\ $$$$\int{dvdt}={at}\Rightarrow{v}={u}+{at}…\left({i}\right) \\ $$$${s}=\int{vdt}={ut}+\frac{{at}^{\mathrm{2}} }{\mathrm{2}}…\left({ii}\right) \\ $$$${v}^{\mathrm{2}} =\left({u}+{at}\right)^{\mathrm{2}} ={u}^{\mathrm{2}} +{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{2}{uat}={u}^{\mathrm{2}} +\mathrm{2}{a}\left(\frac{{at}^{\mathrm{2}} }{\mathrm{2}}+{ut}\right) \\ $$$$\Rightarrow{v}^{\mathrm{2}} ={u}^{\mathrm{2}} +\mathrm{2}{as}…\left({iii}\right) \\ $$