Question

Make s, u subject of the formula

`v^(2) = u {}^(2) + 2as`
​

Answer 1

`\boxed{u = \pm \sqrt{ {v}^(2) - 2as}}`

`\boxed{s = \frac{ {v}^(2) - {u}^(2) }{2a}}`

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

s =
`(v^2-u^2)/(2a)`

Explanation:

Given

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

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

`(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 )

±
`√(v^2-2as)` = u