# SUVAT Equations

Here’s a good one to take home to show the kids, makes learning Algebra fun!

This is a short set of equations, often written using the variable names s, u, v, a, and t, standing for displacement, initial velocity, final velocity, acceleration, and time respectively. SOMETHING OF NOTE: these only work to describe situations with uniform (i.e. constant) acceleration. For the more general case of an accelerating acceleration, you gotta throw in some salt, some pepper, and a pinch of Calculus.

It all starts off by defining average velocity of a body by it’s displacement $v_{average} = \frac{s}{t}$ which reads “the average velocity is the change in displacement over time” – of course, if you know your maths, you know you can define an Arithmetic Average of our velocity here as $v_{average}=\frac{v+u}{2}$ so $s=v_{average}t=(\frac{v+u}{2})t$

Now we can define acceleration similarly, by saying $a = \frac{v-u}{t}$ which reads “acceleration is change in velocity over time”.

Rearrange that to find $v=u+at$ –> obviously, you can also get $t=\frac{v-u}{a}$ from this.

Do some substitution, and find that $s=(\frac{v+u}{2})t=(\frac{u+at+u}{2})t=ut+\frac{1}{2}at^2$ —> solving for u gives $s = vt - \frac{1}{2}at^2$ instead. Handy innit?

Let’s solve for t now (oh, I already did), and sub that into our displacement equation: $s =\frac{u+v}{2}t=\frac{u+v}{2}\frac{v-u}{a} = \frac{(u+v)(v-u)}{2a}$ and rearrange it, to find $v^2=u^2+2as$

And that’s all there is to it! You can solve basic kinetics problems now! Hooray!

I’ll give a brief summary of the equations I guess, I’ve got time. So in no particular order:

• $v=u+at$
• $v^2=u^2+2as$
• $s = (\frac{u+v}{2})t$
• $s = ut + \frac{1}{2}at^2$
• $s = vt - \frac{1}{2}at^2$
• $a = \frac{v-u}{t}$

With these, you’ve got what it takes to do some damage – just don’t forget which way is pointing up! Good luck my friend.

OH! I should also mention now that you’re here, that there are analogous equations to describe Rotational situations, with the variables being replaced by angular analogues. You’ll bump into them, I’m sure.

Au revoir!