# Differentiate the equation for linear motion

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.

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It's not called an exponential curve, it's called a parabola. Instead of calling it a line plus parabola, it might be prudent to complete the square and see it as simply a parabola with a specified minimum/maximum height and vertex. Otherwise this looks fine. – anon Sep 23 '11 at 5:51
@anon I think it's a parabola, not a hyperbola. – Olivier Bégassat Sep 23 '11 at 5:54
Bah, there I go screwing up my high school vocabulary. :) – anon Sep 23 '11 at 5:56
Thanks. Things like that I what I want to learn/relearn. :) And hey, I know that "completing the square". :) Edit time. I'll need to learn/relearn that hyperbola stuff. Probably learned it in High School, but, well, forgot. One reason I'm relearning intermediate math. :) Parabola, check. – Narf the Mouse Sep 23 '11 at 5:58
I don't understand what you're looking for in Edit3. However, in the third point, there's an error: you have to subtract the same quantity in both side, then is $-s_0$ also in the right side. And why you repeat the same expression in the end? – Immanuel Weihnacht Sep 23 '11 at 7:18

Acceleration is $s''$, so $s'' = a$ (=constant)
Integrating this gives the equation for the velocity as s' = $v_0$ + $a\times t$ where $v_0$ is the velocity at time 0
Integrating once more gives $s = s_0 + v_0 t + \frac{1}{2} a t^2$
This curve is the parabola passing through $(0,t)$ and with gradient $v_0$ at $t=0$.