Question

Make s, u subject of the formulaTexFormula1" title="v^{2} = u {}^{2} + 2as" alt="v^{2} = u {}^{2} + 2as" align="absmiddle" class="latex-formula">
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Answer 1

Answer:

$\boxed{u = \pm \sqrt{ {v}^{2} - 2as}}$

$\boxed{s = \frac{ {v}^{2} - {u}^{2} }{2a}}$

Step-by-step explanation:

•Making u subject for formula v² = u² + 2as:

$Solve \: for \: u: \\ = > {v}^{2} = {u}^{2} +2as \\ \\ {v}^{2} = {u}^{2} + 2as \: is \: equivalent \: to \: {u}^{2} + 2as = {v}^{2} : \\ = > {u}^{2} + 2as = {v}^{2} \\ \\ Subtract \: 2as \: from \: both \: sides: \\ {u}^{2} = {v}^{2} - 2as \\ \\ Take \: the \: square \: root \: of \: both \: sides: \\ = > u= \sqrt{ {v}^{2} - 2as } \: \: \: \: or \: \: \: \: u=- \sqrt{ {v}^{2} - 2as} \\ u = \pm \sqrt{ {v}^{2} - 2as}$

•Making s subject for formula v² = u² + 2as:

$Solve \: for \: s: \\ = > {v}^{2} =2as+ {u}^{2} \\ \\ {v}^{2} =2as+ {u}^{2} is equivalent to 2as+ {u}^{2} = {v}^{2}: = > 2as+ {u}^{2} = {v}^{2} \\ \\ Subtract \: {u}^{2} \: from \: both \: sides: \\ = > 2as= {v}^{2} - {u}^{2} \\ \\ Divide \: both \: sides \: by \: 2 a: \\ = > s = \frac{ {v}^{2} }{2a} - \frac{ {u}^{2} }{2a} \\ = > s = \frac{ {v}^{2} - {u}^{2} }{2a}$

Answer 2

Answer:

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

Step-by-step explanation:

Given

v² = u² + 2as ( subtract u² from both sides )

v² - u² = 2as ( divide both sides by 2a )

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

Given

v² = u² + 2as ( subtract 2as from both sides )

v² - 2as = u² ( take the square root of both sides )

± $\sqrt{v^2-2as}$ = u