Help with the Separation of Variables and Integration

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The discussion focuses on the challenges of applying separation of variables and integration in a physics context involving changing velocity and forces. Participants clarify the need to separate variables correctly and integrate with respect to the appropriate bounds, particularly emphasizing the distinction between constants and variables. The conversation highlights the importance of using LaTeX for clear mathematical communication and the application of the chain rule in differentiation. Ultimately, the goal is to derive an ordinary differential equation (ODE) that can be solved to find the change in position (∆x) based on the changing velocity. The thread concludes with guidance on evaluating the final expression for position after integrating.
  • #61
\begin{equation}
\ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x
\end{equation}
 
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  • #62
silento said:
\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.
 
  • #63
\left|\frac{u_f}{u_o}\right| = e^{-\frac{2B}{m}x}
 
  • #64
am I typing something wrong?
 
  • #65
silento said:
am I typing something wrong?
Delimiters? And you can lose the absolute value bars. The RHS is always positive.
 
  • #66
\begin{equation}\left| \frac{u_f}{u_o} \right| = e^{-\frac{2B}{m}x}\end{equation}
 
  • #67
silento said:
\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.
 
  • #68
\begin{equation}u_f = u_o \cdot e^{-\frac{2B}{m}x}\end{equation}
 
  • #69
silento said:
\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##.
 
  • #70
erobz said:
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.
 
  • #71
\begin{equation}
u= a - B \times (54)^2
\end{equation}
 
  • #72
thats uf, a represents the constant
 
  • #73
ui is just a since v=0
 
  • #74
silento said:
\begin{equation}
uf
= a - B \times (54)^2
\end{equation}
That’s just ##u_f##.don’t forget ##u_o##.
 
  • #75
I think you can figure it out from here.
 
  • #76
wait hold on its the other way around
 

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