Help with the Separation of Variables and Integration

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SUMMARY

The discussion focuses on solving a differential equation involving the separation of variables and integration in the context of physics. Participants clarify the relationship between velocity (v), position (x), and forces, specifically addressing the drag force and its non-constant nature. The equation discussed is of the form $$\varphi - \beta v^2 = m v\frac{dv}{dx}$$, with a substitution of $$u = \varphi - \beta v^2$$ to facilitate integration. The final goal is to derive the change in position (∆x) as velocity transitions from 120 m/s to 0 m/s.

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  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with the Chain Rule in calculus
  • Knowledge of basic physics concepts, including forces and motion
  • Proficiency in LaTeX for mathematical formatting
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  • Study the process of solving ordinary differential equations (ODEs) in physics
  • Learn about the Chain Rule and its applications in differentiation
  • Explore the concept of drag force and its impact on motion
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Students and professionals in physics, engineering, and mathematics who are working on problems involving differential equations and motion analysis.

  • #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.
 
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  • #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.
 
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  • #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
 
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  • #74
silento said:
\begin{equation}
uf
= a - B \times (54)^2
\end{equation}
That’s just ##u_f##.don’t forget ##u_o##.
 
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  • #75
I think you can figure it out from here.
 
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  • #76
wait hold on its the other way around
 

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