# Deriving the constant acceleration formulas using functional notation.

#### theo tree

Hi,

So I was trying to develop a more in-depth understanding of the constant acceleration equations, especially v^2 = u^2+2as.

His logic goes like this.

1: Acceleration is the derivative of velocity w.r.t. time.

a = dv/dt

2, Chain rule:

a = dv/ds * ds/dt

3, as ds/dt = v we have:

a = v * dv/ds

4, Times by ds

a ds = v * dv

5, integrating:

integral of a w.r.t. s = integral of v w.r.t. v

6, at t = 0, s = 0 and v = u (initial velocity) so

as - a0 = 1/2 * v^2 - 1/2 * u^2

2as = v^2 - u^2

v^2 = u^2 + 2as

All well and good, right?

Except that in stage 4 he multiplies by "ds". I know that this is common practice amongst all you calculus pros, but strictly speaking "dv/ds" is a limit, not a fraction, right? So you can't treat ds like it was a denominator.

To better understand what is actually going on here, I decided to try and write an analogous argument using functional notation. And that's when I got stuck. What'cha reckon? Here are my current thoughts:

a and u are constants (u is initial velocity - I think that might be British notation?).

t is the independent variable.

v is a function, f, of t, so that f(t) = u + at

s is a function, g, of t, so that g(t) = ut + 1/2 * a * t^2

v can also be written as a function, h, of s, so that h(s) = Dt s (Differentiate s wrt. t)

... so far so good?

We then have:

fÂ´(t) = a and
gÂ´(t) = u + at but I can't figure out what
hÂ´(s) = ?

a = v * dv/ds

In functional notation this would be:

a = v * h'(s) ... right?

So I guess that h'(s) must equal a/v, but I can't explain that at all.

Anyone got any ideas?

Thanks!

Theo

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#### skipjack

Forum Staff
So you can't treat ds like it was a denominator.
The isolated "ds" is how you indicate that you're integrating with respect to s. Because of the chain rule, you can proceed as though you're dealing with fractions, even though you're not.

The video uses subscripts for "initial" and "final". It doesn't do that for the time variable, and it's clearer not to do it for the other variables.

#### skeeter

Math Team
$a(t) = v'(t) = k$ where $k$ is a constant

integrating w/respect to time yields the equation for velocity under uniform acceleration ...

$v(t) = kt+C_1$

$v(0) = u \implies C_1 = u$

$v(t) = s'(t) = kt + u$

integrating w/respect to time again yields the position equation ...

$s(t) = \dfrac{kt^2}{2} + ut + C_2$

$s(0) = C_2 = s_0$

$s(t) = \dfrac{kt^2}{2} + ut + s(0)$

$s(t) - s_0 = ut + \dfrac{kt^2}{2}$

which leads to the familiar displacement equation for uniform (constant) acceleration ...

$\Delta s = ut + \dfrac{1}{2}at^2$

$v = u + at \implies t = \dfrac{v-u}{a}$

substituting for $t$ in the displacement equation ...

$\Delta s = \dfrac{u(v-u)}{a} + \dfrac{a}{2} \cdot \left(\dfrac{v-u}{a}\right)^2$

$\Delta s = \dfrac{uv-u^2}{a} + \dfrac{(v-u)^2}{2a}$

common denominator is $2a$ ...

$\Delta s = \dfrac{2(uv-u^2) + (v-u)^2}{2a}$

$2a \Delta s = 2uv - 2u^2 + v^2 - 2uv + u^2$

combine like terms ...

$2a \Delta s = v^2 - u^2 \implies v^2 = u^2 + 2a \Delta s$

#### theo tree

Hi,

Thanks for the replies. Especially to skeeter for the neat derivation of s=ut+1/2ut^2.

I'm still stuck on figuring out exactly what is going on in (4.) tho...

Skeeter, your proof of v^2 = u^2 + 2as is super neat, but not analagous to the one in the video. I'm really after the same proof, just written in functional notation, so as to clear up this whole "times by ds" business.

What is really being integrated there?

a = v * dv/ds

Do we integrate both sides with repect to s? If so, do we need to write v as a function of s?

Or do we first divide through by dv/ds and then integrate a over dv/ds wrt. s?

I'm looking for clarity wrt. my integration Thanks for the help!

#### skipjack

Forum Staff
Both sides are integrated with respect to s. If dv/ds is defined, integrating v * dv/ds with respect to s is the same as integrating v with respect to v, so the integration can be done without already knowing how to express v as a function of s.

#### theo tree

Awesome! Thank you 