Respuesta :

Itzisabella

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

TheAnimeGirl

Answer:

[tex] \boxed{ \boxed{ \orange{ \sf{see \: below}}}} [/tex]

Explanation:

[tex] { \underline{ \underline \bold{ \sf{ \blue{ {question \: }}}}}}[/tex]

[tex] \sf{ \bold{ \underline{ \: prove \: that \: {v}^{2} = {u}^{2} + 2as}}}[/tex]

[tex] \underline{ \underline{ { \bold{ \sf{ \purple{solution} }}}}}[/tex]

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 ,

[tex] \sf{v = u + at}[/tex] ⇒first equation of motion ( i )

[tex] \sf{s = \frac{u + v}{2} \times t}[/tex] ⇒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 }

[tex] \sf{s = \frac{u + v}{2} \times \frac{v - u}{a} }[/tex]

[tex] \sf{or \: s = \frac{ {v}^{2} - {u}^{2} }{2a} }[/tex]

[tex] \sf{or \: 2as = {v}^{2} - {u}^{2} }[/tex]

[tex] \sf{ \boxed{ \bold{ \:  ∴ \: \: \: {v}^{2} = {u}^{2} + 2as}}}[/tex] ⇒ forth equation of motion

Hope I helped!

Best regards!!