Help with the Separation of Variables and Integration

[silento]

but there is two answers. a + and a - since we were taking the square root

 

[silento]

?

 

[erobz]

The equation you need to solve is:

$$\frac{- 2 \beta}{m} u = \frac{du}{dx} $$

Separate variable's and integrate.

And if you don't reply in Latex math formatting I'm calling it quits with a clear conscience...
See: LaTeX Guide

 

[silento]

\begin{equation}
\ln|u| = -\frac{2B}{m}x
\end{equation}

 

[erobz]

\begin{equation}
\ln|u| = -\frac{2B}{m}x
\end{equation}

Integrate $u$ with limits $u_f, u_o$. You can leave the $x$ the way it is assuming $x_o=0$. Otherwise good.

 

[silento]

so let me think about this. u is velocity from 54 m/s to 0 m/s. x_f is what I'm solving for. Would I only integrate the LHS?

 

[erobz]

so let me think about this. u is velocity from 54 m/s to 0 m/s. x_f is what I'm solving for. Would I only integrate the LHS?

$u$ is not $v$. Go back and look at what $u$ is…

 

[erobz]

The basic equation is fine. But it’s missing a constant of integration. That is why I suggest you “re do it” integrating it as a definite integral between limits $u_f$ and $u_o$.

 

[silento]

\begin{equation}
\ln|u_f| - \ln|u_o| = -\frac{2B}{m}x
\end{equation}

 

[erobz]

\begin{equation}
\ln|u_f| - \ln|u_o| = -\frac{2B}{m}x
\end{equation}

Now combine the logs

 

[silento]

\begin{equation}
\ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x
\end{equation}

 

[erobz]

\begin{equation}
\ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x
\end{equation}

exponentiate both sides, and move $u_o$ over to the right hand side.

 

[silento]

\left|\frac{u_f}{u_o}\right| = e^{-\frac{2B}{m}x}

 

[silento]

am I typing something wrong?

 

[erobz]

am I typing something wrong?

Delimiters? And you can lose the absolute value bars. The RHS is always positive.

 

[silento]

\begin{equation}\left| \frac{u_f}{u_o} \right| = e^{-\frac{2B}{m}x}\end{equation}

 

[erobz]

\begin{equation}\left| \frac{u_f}{u_o} \right| = e^{-\frac{2B}{m}x}\end{equation}

The absolute value bars are no longer necessary. The RHS is always positive. Move $u_o$ over.

 

[silento]

\begin{equation}u_f = u_o \cdot e^{-\frac{2B}{m}x}\end{equation}

 

[erobz]

\begin{equation}u_f = u_o \cdot e^{-\frac{2B}{m}x}\end{equation}

Now evaluate $u_f$ and $u_o$ using the corresponding $v_f$ and $v_o$. Make sure you are evaluating $u$. I repeat $u$ is not $v$.

 

[erobz]

Now evaluate $u_f$ and $u_o$ using the corresponding $v_f$ and $v_o$. Make sure you are evaluating $u$. I repeat $u$ is not $v$.

Oh I just realized you are trying to solve for $x$, not $v$. Anyhow just finish it out, the solve for $x$. I realized this is undoing a few steps. My bad.

 

[silento]

\begin{equation}
u= a - B \times (54)^2
\end{equation}

 

[silento]

thats u_f, a represents the constant

 

[silento]

u_i is just a since v=0

 

[erobz]

\begin{equation}
u_f
= a - B \times (54)^2
\end{equation}

That’s just $u_f$.don’t forget $u_o$.

 

[erobz]

I think you can figure it out from here.

 

[silento]

wait hold on its the other way around