Derive Formula for Final Velocity (v2) Without Time

[NeedBetterName]

Homework Statement

Derive the formula for final velocity, $v_2$, without time included in said formula.

$\overrightarrow{a}$: Known

$\overrightarrow{s}$: Known

$v_1$: Known

$\Delta t$: Unknown (can be calculated, but defeats the purpose)

$\overrightarrow{v_2}$: Unknown and to be calculated

Homework Equations

$$\begin{align*}<br /> \Delta t &= \frac{\overrightarrow{s}}{\overrightarrow{\Delta v}} \\ \\<br /> \overrightarrow{a} &= \frac{\overrightarrow{\Delta v}}{\Delta t}<br /> \end{align*}$$

The Attempt at a Solution

Since I've always disliked memorizing formulas, I just figure out how they're derived and then if I forget the formula, I can get it regardless; so I went about trying to get the formula as follows:
$$\begin{align*}<br /> \overrightarrow{a} &= \frac{\overrightarrow{\Delta v}}{\Delta t} \\<br /> &= \frac{\overrightarrow{\Delta v}}{\frac{\Delta \overrightarrow{s}}{\Delta \overrightarrow{v}}} \\<br /> &= \frac{\Delta \overrightarrow{v}^2}{\Delta \overrightarrow{s}} \\<br /> &= \frac{(v_2 - v_1)^2}{\Delta \overrightarrow{s}} \\<br /> \overrightarrow{a} \Delta \overrightarrow{s} &= (v_2 - v_1)^2 \\<br /> &= v_2^2 - 2v_2v_1 - v_1^2 \\ \\<br /> v_2^2 &= -2v_2v_1 - v_1^2 + \overrightarrow{a} \Delta \overrightarrow{s}<br /> <br /> \end{align*}$$

I ended up with a result that different than the formula taught in my class. I can't remember the formula taught in class, as I wasn't actually there for the lesson, but I vaguely remember what the formula was like, when I saw it later. I foolishly forgot to write it down, so now I'm not too sure, but I think it was similar to [itex]\v_2^2 = v_1^2 + \overrightarrow{a}\Delta \overrightarrow{s}[\itex].<br /> <br /> <b>4. What I really want to know</b><br /> I basically need to know 4(ish) things: Is my derived formula okay? If not, where'd I go wrong? What's the formula similar to what I listed above (or is it correct; this question also isn't too big of a deal since I can get it tomorrow)? And if the second formula is correct (or the correct, similar one has been provided), how is it derived?<br /> <br /> <b>EDIT:</b> I remember the formula in class, which is $v_2^2 = v_1^2 + 2 \overrightarrow{a} \Delta \overrightarrow{s}$. How is this derived?[/itex]

 

[tms]

$$\begin{align*}<br /> \overrightarrow{a} &= \frac{\overrightarrow{\Delta v}}{\Delta t} \\<br /> &= \frac{\overrightarrow{\Delta v}}{\frac{\overrightarrow{v}}{\Delta t}}<br /> \end{align*}$$
The denominator in the second line doesn't look right. Instead, start by changing $\Delta v$ to $v_f - v_i$, and then solve for $t$.

 

[NeedBetterName]

$$\begin{align*}<br /> \overrightarrow{a} &= \frac{\overrightarrow{\Delta v}}{\Delta t} \\<br /> &= \frac{\overrightarrow{\Delta v}}{\frac{\overrightarrow{v}}{\Delta t}}<br /> \end{align*}$$
The denominator in the second line doesn't look right. Instead, start by changing $\Delta v$ to $v_f - v_i$, and then solve for $t$.

That was actually just a typing mistake, which I just fixed. The rest of my work went on as if it was $$\overrightarrow{a} = \frac{\Delta \overrightarrow{v}}{\frac{\Delta \overrightarrow{s}}{\Delta \overrightarrow{v}}}$$

And by solving for time, as mentioned, defeats the purpose of the formula. Having it as one single formula where I just input all values and solve immediately makes it quicker and cleaner.

 

[tms]

And by solving for time, as mentioned, defeats the purpose of the formula.

That's just the first step.

Having it as one single formula where I just input all values and solve immediately makes it quicker and cleaner.

But your initial result is wrong.

BTW, you can use \vec for vectors, which is easier to type than \overrightarrow.

 

[NeedBetterName]

I found out where the equation is derived from when I noticed that one of the formulas looked a lot like the area of a trapezoid formula, which lead me to a velocity-time graph and I was able to derive from there.

But your initial result is wrong.

The reason I asked was to check if it worked, and if not, check why it didn't work. Although it doesn't really matter much anymore, since I found out how the correct one is derived.

BTW, you can use \vec for vectors, which is easier to type than \overrightarrow.

Ah, much better. I was recently typing LaTeX in an online LaTeX equation editor, and \overrightarrow looked better than \vec, but on Physics Forums, \vec appears nicer, and easier to write too.

 

[SammyS]

Homework Statement

Derive the formula for final velocity, $v_2$, without time included in said formula.

$\overrightarrow{a}$: Known

$\overrightarrow{s}$: Known

$v_1$: Known

$\Delta t$: Unknown (can be calculated, but defeats the purpose)

$\overrightarrow{v_2}$: Unknown and to be calculated

Homework Equations

$$\begin{align*}<br /> \Delta t &= \frac{\overrightarrow{s}}{\overrightarrow{\Delta v}} \\ \\<br /> \overrightarrow{a} &= \frac{\overrightarrow{\Delta v}}{\Delta t}<br /> \end{align*}$$

The Attempt at a Solution

Since I've always disliked memorizing formulas, I just figure out how they're derived and then if I forget the formula, I can get it regardless; so I went about trying to get the formula as follows:$$\begin{align*}<br /> \overrightarrow{a} &= \frac{\overrightarrow{\Delta v}}{\Delta t} \\<br /> &= \frac{\overrightarrow{\Delta v}}{\frac{\Delta \overrightarrow{s}}{\Delta \overrightarrow{v}}} \\<br /> &= \frac{\Delta \overrightarrow{v}^2}{\Delta \overrightarrow{s}} \\<br /> &= \frac{(v_2 - v_1)^2}{\Delta \overrightarrow{s}} \\<br /> \overrightarrow{a} \Delta \overrightarrow{s} &= (v_2 - v_1)^2 \\<br /> &= v_2^2 - 2v_2v_1 - v_1^2 \\ \\<br /> v_2^2 &= -2v_2v_1 - v_1^2 + \overrightarrow{a} \Delta \overrightarrow{s}<br /> <br /> \end{align*}$$
I ended up with a result that different than the formula taught in my class. I can't remember the formula taught in class, as I wasn't actually there for the lesson, but I vaguely remember what the formula was like, when I saw it later. I foolishly forgot to write it down, so now I'm not too sure, but I think it was similar to [itex]\v_2^2 = v_1^2 + \overrightarrow{a}\Delta \overrightarrow{s}[\itex].<br /> <br /> <b>4. What I really want to know</b><br /> I basically need to know 4(ish) things: Is my derived formula okay? If not, where'd I go wrong? What's the formula similar to what I listed above (or is it correct; this question also isn't too big of a deal since I can get it tomorrow)? And if the second formula is correct (or the correct, similar one has been provided), how is it derived?<br /> <br /> <b>EDIT:</b> I remember the formula in class, which is $v_2^2 = v_1^2 + 2 \overrightarrow{a} \Delta \overrightarrow{s}$. How is this derived?[/itex]

[itex] It seems that $\ v_2^2 = v_1^2 + 2 \vec{a}\cdot \Delta \vec{s}\$ should be derived in your textbook !<br /> <br /> This (erroneous) equation is causing at least some of the problem.<div style="margin-left: 20px">This is wrong → $\displaystyle \Delta t = \frac{\vec{s}}{\vec{\Delta v}}\ \$<br />​</div>(Besides the fact that division of a vector by a vector is not defined:)<br /> <br /> The correct equation comes from<div style="margin-left: 20px">$\displaystyle \vec{v}_\text{average}=\frac{\Delta \vec{s}}{\Delta t}$​</div>So that $\displaystyle \ \ \vec{v}_\text{average}\,\Delta t= \Delta \vec{s}\ .$<br /> <br /> For uniform acceleration, $\displaystyle \ \ \vec{v}_\text{average} = (\vec{v}_1 + \vec{v}_2)/2\ .$<br /> <br /> Also, $\displaystyle \ \ \vec{a}_\text{average}=\frac{\Delta \vec{v}}{\Delta t}\ .\ \$ Furthermore, for uniform acceleration, $\displaystyle \ \ \vec{a} = \vec{a}_\text{average}\ .$<br /> <br /> Now, consider $\displaystyle \ \ \vec{v}_\text{average}\cdot(\Delta \vec{v})\ .$[/itex]