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How shall I derive, $v^2=u^2+2as$ using the equation $F=ma$?

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  • $\begingroup$ How people could understand a question based on undefined symbols? I am just writing this comment. Other people could downvote the question. $\endgroup$ – GiorgioP Apr 10 at 16:41
  • $\begingroup$ I have used tag Newtonian mechanics, kinematics. Mentioned '3rd equation of motion' in the title... And u are asking how to understand the symbols... Wow! 😊 $\endgroup$ – Swayam Jha Apr 10 at 16:50
  • $\begingroup$ Can u tell me in physics, F=ma stands for what other than force, acceleration and mass ? $\endgroup$ – Swayam Jha Apr 10 at 16:51
  • $\begingroup$ The symbols are clear. $\endgroup$ – Andrew Steane Apr 10 at 19:46
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    $\begingroup$ @AndrewSteane No problem for $F,m,a$, but why should I know what $u$ is and what relation has with $v$? In a scientific paper, the first request by any referee would be to clarify the meaning of symbols. I also would like to know how many textbooks call that equation 3rd equation of motion. After years of physics, this is the first time I see that name. $\endgroup$ – GiorgioP Apr 10 at 22:30
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$F=ma=m\frac{dv}{dt}$

by chain rule

$m\frac{dv}{dx}\frac{dx}{dt}=m\frac{dv}{dx}v=F$

Rearrangment yield $\int Fdx=m\int vdv$, this derivation implicitly imply force is constant then $F=ma$

Integration gives $max=m(\frac{v^2}{2}-\frac{u^2}{2})$, cancel common factor mass rearrange gives final result:

$2ax=v^2-u^2$

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  • $\begingroup$ But we can't use this method to derive the equation of motions for non uniform acceleration.... Thus, can u provide a general method which can even use for the derivation of non uniform acceleration motion...? $\endgroup$ – Swayam Jha Apr 10 at 16:33
  • $\begingroup$ You can't derive your equation for non-uniform motion because it is not true unless $a$ is constant. $\endgroup$ – mike stone Apr 10 at 19:52
  • $\begingroup$ Why do you call it the "third equation of motion"? This makes it sound like some general principle, rather than an equation with very restricted application. $\endgroup$ – mike stone Apr 10 at 19:56
  • $\begingroup$ Sorry for wrong terminology. Actually, in my highschool textbook it's referred as 'equations of motion. $\endgroup$ – Swayam Jha Apr 11 at 2:23
  • $\begingroup$ And my teacher too use this terminology.. So may you please tell me what would be the correct vocabulary? $\endgroup$ – Swayam Jha Apr 11 at 2:23
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Increase of kinetic energy is equal to the work done, which is given by the force multiplied by the distance moved in the direction of the force. In other words:

(final kinetic energy) = (initial kinetic energy) + $F s$

which is $$ \frac{1}{2} m v^2 = \frac{1}{2} m u^2 + m a s $$ Now you can multiply by 2 and divide by $m$. The advantage of this method is that it shows the physical meaning of this result, which is contained in the first version in words. Also, that first version is all you need to remember, because the final result follows immediately so you can derive it whenever you need it.

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  • $\begingroup$ Thanks alot for the variation.... Worth noting that there are multiple ways to reach a single equation.... $\endgroup$ – Swayam Jha Apr 10 at 16:21
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Let the force(F) act on the body and displace it by 'S'. Then work done by this force equals change in Kinetic energy. Let initial velocity be u and final velocity be v. Now,

Work done by F = $\Delta$ KE

F.S = $\frac12$ mv$^2$ - $\frac12$ mu$^2$ ... i

Also from Newton's second law : F = ma

Substituting this result in i we get,

ma.S = $\frac12$ mv$^2$ - $\frac12$ mu$^2$

Multiplying both sides by 2 and cancelling m on both sides

2a.S = v$^2$ - u$^2$

Thus

v$^2$ = u$^2$ + 2aS [QED]

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$$v=u+at$$$$v^2=u^2+2uat+a^2t^2$$$$v^2=u^2+2a(ut+\frac{1}{2}at^2)$$$$v^2=u^2+2as$$ Quite starightforward.

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