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In the following equation $$a = \frac{v^2 − u^2 } {2s} ,$$ where $v$ is the final velocity, $u$ is the initial velocity, and $s$ is displacement. Why is velocity squared and displacement multiplied by 2?

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closed as off-topic by Aaron Stevens, Jon Custer, GiorgioP, stafusa, ZeroTheHero Mar 8 at 17:44

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    $\begingroup$ Have you tried to derive this formula? Start from $v=u+a t$ and $s=u t+\frac{1}{2}a t^2$. $\endgroup$ – G. Smith Mar 7 at 18:14
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    $\begingroup$ In these simpler equations for how velocity and displacement depend on time, do you understand why the second term in the second one has $t$ squared and a factor of 1/2? These two things, plus algebra, lead to the velocities being squared in your equation, and the displacement being multiplied by 2. $\endgroup$ – G. Smith Mar 7 at 18:25
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Mar 7 at 18:47
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The kinematic equation for constant acceleration $$a = \dfrac{(v^2 − u^2 )} {2s}$$ can be changed to the following equation where $m$ is the mass of the object and $F$ is the force acting on the object.

$$\frac 12 m v^2 - \frac 12 m u^2 = ma\, s = F\, s$$

The left hand side is the change in the kinetic energy of the object and the right hand side is the work done on the object by an external force.

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  • $\begingroup$ I don't think this is a worthwhile direction for this kinematics question. G Smith's comment is far more likely the right way to go. $\endgroup$ – Kyle Kanos Mar 8 at 11:10

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