Question

12. Prove that v' = u? + 2as.​

Answer 1

Correction:-

Prove that v² = u² + 2as .

Solution:-

From first equation of motion,

v = u + at

=> at = v - u

=> t = v - u / a

From the second equation of motion, we have

s = ut + 1/2at²

Putting the value of t in above equation , we get:

s = u ( v - u /a ) + 1/2 a ( v - u/a )²

s = uv - u² / a + a( v² + u² - 2uv / 2a²)

s = uv - u² / a + v² + u² - 2uv / 2a

s = 2uv - 2u² + v² + u² - 2uv / 2a

2as = v² - u²

Answer 2

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Step-by-step explanation:

`{ \underline{ \underline \bold{ \sf{ \blue{ {question \: }}}}}}`

`\sf{ \bold{ \underline{ \: prove \: that \: {v}^(2) = {u}^(2) + 2as}}}`

`\underline{ \underline{ { \bold{ \sf{ \purple{solution} }}}}}`

Let us assume a body moving with an initial velocity ' u '. Let it's final velocity be 'v' after a time 't' and the distance travelled by the body be 's'. We already have ,

`\sf{v = u + at}` ⇒first equation of motion ( i )

`\sf{s = (u + v)/(2) * t}` ⇒second equation of motion ( ii )

Putting the value of t from ( i ) in the equation ( ii )

{ v = u + at

or , at = v - u

or, t = v-u / a }

`\sf{s = (u + v)/(2) * (v - u)/(a) }`

`\sf{or \: s = \frac{ {v}^(2) - {u}^(2) }{2a} }`

`\sf{or \: 2as = {v}^(2) - {u}^(2) }`

`\sf{ \boxed{ \bold{ \:  ∴ \: \: \: {v}^(2) = {u}^(2) + 2as}}}` ⇒ forth equation of motion

Hope I helped!

Best regards!!