Learning calculus in my uncopious spare time. Got to this far and decided to post this, Just to check if I've got this right - Tell me if it's wrong and where. :) (I don't have a college professor to do that)
Nothin' up my sleeve, or; done right here, in the post area, with no calculators of any kind:
$s_1 = s_0 + vt + \frac{1}{2} at^2$ .
(Drop s0, as it doesn't contribute towards the curve of the derivative, just moves the entire curve up the $y$-axis; change $vt^1$ to $1vt^0$ to $v*1$ to $v$; change $\frac{1}{2}at^2$ to $2 \times \frac{1}{2} \times at$ to $at$) ->
s'_1 = v + at ->
(drop $v$, for the same reason for this derivative; change $at^1$ to $1at^0$ to $a \times 1$ to $a$) ->
s''_1 = a.
s''_1 = \text{constant}.
s'_1 = \text{line}.
$s_1$ = Curve mapped by a line + (an exponential (^2) curve, halved), shifted up the $y$-axis by $s_0$.
So, what did I get wrong and how? Thanks.
Edit: Changed "$vt^0$" to "$1vt^0$" and "$at^0$" to "$1at^0$".
Edit2: ...Maybe I need to review completing the square. I know how to unfold it, but folding it is escaping me at the moment. OTOH, it is 11:05 PM.
Edit3: I'd like to take this moment to thank Khan Acamedy.
$s_0 + vt + \frac{1}{2} at^2 = 0$
$\frac{1}{2} at^2 + vt + s_0 = 0$
$\frac{1}{2} at^2 + vt + s_0 - s_0 = 0 + s_0$
$\frac{1}{2} at^2 + vt = s_0$
$\frac{1}{2} at^2 + vt = s_0, b = \frac{v}{2}$
$\frac{1}{2} at^2 + vt + b^2 = s_0 + b^2, b = \frac{v}{2}$
...Ok, how do you handle that fraction? Er, the one in front of the "$at^2$".
$(\frac{1}{2} at + b)^2 = s_0 + b^2, b = \frac{v}{2}$
$(\frac{1}{2} at + b) * (\frac{1}{2} at + b) = s_0 + b^2, b = \frac{v}{2}$
$\frac{1}{4} at^2 + \frac{1}{2} atb + \frac{1}{2} atb + b^2 = s_0 + b^2, b = \frac{v}{2}$
Nope.
$\frac{1}{2}(at + b)^2 = s_0 + b^2, b = \frac{v}{2}$
$\frac{1}{2}((at + b) * (at + b)) = s_0 + b^2, b = \frac{v}{2}$
$\frac{1}{2}(2at^2 + atb + atb + b^2), b = \frac{v}{2}$
$(at^2 + 2abt + b^2), b = \frac{v}{2}$
...Yeah, I need more math.