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}
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}