Where do the suvat equations come from?

Most of the suvat equations are pretty easy to derive, as soon as you realise acceleration ($a$, assumed constant) is the derivative of velocity ($v$) with respect to time, and velocity is the derivative of position ($s$), also with respect to time. For example:

$ a = \diff{v}{t}$
$ \int_0^t a \d{T} = \int_u^v dV$
$ at = v - u$, or $v = u + at$.

You can also, if you're into the velocity-time graph, see that drawing a straight line through $(0,u)$ with gradient $a$ gives an equation $v = u+at$.

Similarly, $v = \diff{s}{t}$
$\int_0^t V \d{T} = \int_0^s \d{S}$
$\int_0^t (u + aT)\d{T} = \int_0^s \d{S}$
$ut + \frac 12 at^2 = s$.

This also comes out of the velocity-time graph, although it's less obvious: you're after the area of a trapezium with parallels $u$ and $v$, and height $t$, giving $s = \frac 12 t (u+v) = \frac 12 t(2u + at) = ut + \frac 12 at^2$.

These two are the ones you use most often in C1, although there are three more. (Each of the five equation relates four of the five quantities ($a$, $u$, $v$ $s$ and $t$) and ignores the last).

One of the others, you've already seen -- the trapezium area $s = \frac 12 t (u+v)$ is one that doesn't directly involve the acceleration. You don't need more calculus for this one -- you can get it by eliminating $a$ from the first two equations.

Then there were two. The fourth is the Cinderella, and very rarely gets used: $s = vt - \frac 12 at^2$. Again, this is a simple rearrangement of the second equation -- replace $u$ with $v - at$ and you're there.

The last one, $v^2 = u^2 + 2as$ is a bit of an odd one out -- it's much trickier to derive than the others, but it's tremendously useful.

Obviously, you could get it by eliminating $t$ from the first two equations:
$t = \frac{v-u}{a}$, so
$s =u \frac{v-u}{a} + \frac 12 a \left(\frac{v-u}{a} \right)^2$, and then a load of tedious algebra.

Much more nicely: $v-u = at$, from the first equation; $v + u = \frac{2s}{t}$, from the third; multiplying the two together gives $v^2 - u^2 = at \times \frac{2s}{t} = 2as$. That's pretty.

You can also do this with calculus, but it requires a trick: $a=\diff{v}{t} = \diff{v}{s} \diff{s}{t} = v \diff {v}{s}$. Knowing that, you can double1 and integrate: $\int_0^s 2a \d{S} = \int_u^v 2V \d{V}$

$2as = v^2 - u^2$, as required!


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. for simplicity's sake []


Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter