silento
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\begin{equation}
\ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x
\end{equation}
\ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x
\end{equation}
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.
PREREQUISITESStudents and professionals in physics, engineering, and mathematics who are working on problems involving differential equations and motion analysis.
exponentiate both sides, and move ##u_o## over to the right hand side.silento said:\begin{equation}
\ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x
\end{equation}
Delimiters? And you can lose the absolute value bars. The RHS is always positive.silento said:am I typing something wrong?
The absolute value bars are no longer necessary. The RHS is always positive. Move ##u_o## over.silento said:\begin{equation}\left| \frac{u_f}{u_o} \right| = 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##.silento said:\begin{equation}u_f = u_o \cdot e^{-\frac{2B}{m}x}\end{equation}
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.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##.
That’s just ##u_f##.don’t forget ##u_o##.silento said:\begin{equation}
uf
= a - B \times (54)^2
\end{equation}