# In the equation for acceleration (with known $v$, $u$ and $s$) why velocity is squared and displacement is multiplied by 2? [closed]

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?

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

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• Have you tried to derive this formula? Start from $v=u+a t$ and $s=u t+\frac{1}{2}a t^2$. – G. Smith Mar 7 at 18:14
• 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. – G. Smith Mar 7 at 18:25
• If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Mar 7 at 18:47

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