Question

Proofread. Can you help me edit my proof. I'm trying to trim the fat off acceleration formulas to explain it better.

Answer 1

Acceleration is a change in velocity over time
\[a=\frac{\Delta v}{\Delta t}=\frac{v-v_{0}}{t}\]
Rearranging for velocity
\[v=v_{0}+at\]
Rearranging for time
\[t=\frac{v-v_{0}}{a}\]
For constant acceleration the average velocity is half the difference in velocity, as well as the total distance over time
\[v_{av}=\frac{v_{0}+v}{2}=\frac{x-x_{0}}{t}\]
Rearranging to show x
\[x=x_{0}+(\frac{v_{0}+v)}{2})t\]
Substitute v or t
\[x=x_{0}+(\frac{v_{0}+\textbf{v})}{2})t\hspace{40pt}x=x_{0}+(\frac{v_{0}+v)}{2})\textbf{t}\]
\[x=x_{0}+(\frac{v_{0}+(v_{0}+at)}{2})t\hspace{40pt}x=x_{0}+(\frac{v_{0}+v)}{2})(\frac{v-v_{0}}{a})\]
\[x=x_{0}+\frac{2v_{0}t+at^{2}}{2}\hspace{40pt}x=x_{0}+\frac{v_{0}v-v_{0}^{2}+v^{2}-v_{0}v}{2a}\]
\[x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\hspace{40pt}v^{2}=v_{0}^{2}+2a(x-x_{0})\]

Answer 2

I changed the first line to
Acceleration \(a\) is a change \(\Delta\) in velocity \(v\) from the initial velocity \(v_0\) over time \(t\).

Answer 3

Every equation you used is valid only if acceleration is constant, not just the ones starting with the fourth equation.

Answer 4

Thanks, first line is now
While acceleration is constant, \(a\) is a change \(\Delta\) in velocity \(v\) from the initial velocity \(v_0\) over time \(t\).

Answer 5

nice calc-free approach ^_^ This is annoyingly picky, but Delta t only equals t at t_o = 0 :P I'm a stickler.

Right under "rearranging to find x" there are a few parenthetical mistypes on the numerator is the only thing I can find to fix. Also, maybe if you make all of the velocity terms (v + v_o) to mirror the form (v - v_o) it'll flow better, and you won't have to do that wonky spread out quadratic 2nd up from the bottom on the left.

Last, if you do " \left( and \right) " for parenthesis with fractions

\left( \frac{}{} \right)

, it looks super slick

\[ (\frac{ \textrm{normal parenthesis}}{\textrm{are tiny!!}}) \ \ \ \left( \frac{ \textrm{but these ones are}}{\textrm{super snazzy and profo ~_^}}\right)\]

Awesome job all around though!

Answer 6

@AllTehMaffs I really am looking for spit and polish, so having someone be annoyingly picky is exactly why I posted. I will implement your changes presently.
Thank you so much!

Answer 7

I don't want to be stuck with t0 = 0 for a general formula, thanks @simplephysics. But I also don't want to have delta t in all the formulas. How do I shave off the delta?

Answer 8

I suppose you would have to include a statement at the top specifying that t represents any time interval? \[t = t _{f}-t _{i}\]

Answer 9

I think t0 has to be 0 for your equation to work and make sense, if not, you wouldn't be talking about a point in time, but about what the speed is at a specific interval delta t equal to t.

Answer 10

While acceleration is constant, acceleration \(a\) is a change \(\Delta\) in velocity \(v\), from the initial velocity \(v_0\), over a change in time \(t\) from the initial time \(t_0=0\).
\[a=\frac{\Delta v}{\Delta t}=\frac{v-v_{0}}{t-t_0}=\frac{v-v_{0}}{t}\]
Rearranging for velocity
\[v=v_{0}+at\]
Rearranging for time
\[t=\frac{v-v_{0}}{a}\]
Average velocity is half the difference in velocity, as well as the total distance over time
\[v_{av}=\frac{v+v_{0}}{2}=\frac{x-x_{0}}{t}\]
Rearranging to show x
\[x=x_{0}+\left(\frac{v+v_{0}}{2}\right)t\]
Substitute \(\color{red}{v}\) or \(\color{blue}{t}\)
\[x=x_{0}+\left(\frac{\color{red}{v}+v_0}{2}\right)t\hspace{75pt}x=x_{0}+\left(\frac{v+v_{0}}{2}\right)\color{blue}{t}\]
\[x=x_{0}+\left(\frac{(v_{0}+at)+v_0}{2}\right)t\hspace{40pt}x=x_{0}+\left(\frac{v+v_{0}}{2}\right)\left(\frac{v-v_{0}}{a}\right)\]
\[x=x_{0}+\frac{2v_{0}t+at^{2}}{2}\hspace{75pt}x=x_{0}+\frac{v^2+v_{0}v-v_{0}v-v_0^{2}}{2a}\]
\[x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\hspace{75pt}v^{2}=v_{0}^{2}+2a(x-x_{0})\]

Answer 11

```
While acceleration is constant, acceleration \(a\) is a change \(\Delta\) in velocity \(v\), from the initial velocity \(v_0\), over a change in time \(t\) from the initial time \(t_0=0\).
\[a=\frac{\Delta v}{\Delta t}=\frac{v-v_{0}}{t-t_0}=\frac{v-v_{0}}{t}\]
Rearranging for velocity
\[v=v_{0}+at\]
Rearranging for time
\[t=\frac{v-v_{0}}{a}\]
Average velocity is half the difference in velocity, as well as the total distance over time
\[v_{av}=\frac{v+v_{0}}{2}=\frac{x-x_{0}}{t}\]
Rearranging to show x
\[x=x_{0}+\left(\frac{v+v_{0}}{2}\right)t\]
Substitute \(\color{red}{v}\) or \(\color{blue}{t}\)
\[x=x_{0}+\left(\frac{\color{red}{v}+v_0}{2}\right)t\hspace{75pt}x=x_{0}+\left(\frac{v+v_{0}}{2}\right)\color{blue}{t}\]
\[x=x_{0}+\left(\frac{(v_{0}+at)+v_0}{2}\right)t\hspace{40pt}x=x_{0}+\left(\frac{v+v_{0}}{2}\right)\left(\frac{v-v_{0}}{a}\right)\]
\[x=x_{0}+\frac{2v_{0}t+at^{2}}{2}\hspace{75pt}x=x_{0}+\frac{v^2+v_{0}v-v_{0}v-v_0^{2}}{2a}\]
\[x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\hspace{75pt}v^{2}=v_{0}^{2}+2a(x-x_{0})\]

Answer 12

Should the vectors be bold?

Answer 13

technically yah. Or you could do /vec

\[\vec v \]
the bold is fancier though
\[ \textbf v\]

If you are doing vectors though, you should change your x to something that isn't a cartesian coordinate; s or d something.

Answer 14

I've heard people say vee naught. I thought that meant velocity at something-or-other zero \(v_0\). Is it vee oh \(v_o\)?

Answer 15

It is vee naught, and after posting that I checked all my text books and it's definitely a zero as the subscript, so I'm just crazy. sorry 'bout that ^^

Answer 16

I admire a person who researches their own beliefs, and finds them wanting. I wish I could offer you more medals. Thank you, sincerely, for your help.

Answer 17

I know to derive this in a shorter way.. using vt graph.. if u use vt graph then u can derive them geometrically.. which seems easier cause not much algebra!

Answer 18

@Mashy I'd love to read it!

Answer 19

here it is.. i dunno if i actually made it lengthier :D

Answer 20

I stay true to my word, I loved reading that. @Mashy Does the exercise hold true if acceleration isn't constant?

Answer 21

same lecture in Gilbert Strang's Physics' book

Answer 22

In every physics book. I'm not claiming originality. Just seeking efficiency.

Answer 23

not your algebra manipulation, but the derivative

Answer 24

no.. if acceleration is not a constant, then the graph wouldn't be a straight line!