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 revolves around the separation of variables and integration in the context of physics, specifically dealing with forces and motion. Participants are exploring how to set up and solve a differential equation involving changing velocity and forces acting on an object.
The discussion is active, with participants providing guidance on using substitutions and the chain rule. There is a recognition of the need to clarify definitions and assumptions, particularly regarding the constants and variables involved in the equations.
Some participants mention constraints such as the bounds for velocity and the need to find the change in position (∆x). There is also a reference to the importance of using proper mathematical formatting in the discussion.
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}