Recent content by silento

wait hold on its the other way around

ui is just a since v=0

thats uf, a represents the constant

\begin{equation} u= a - B \times (54)^2 \end{equation}

\begin{equation}u_f = u_o \cdot e^{-\frac{2B}{m}x}\end{equation}

\begin{equation}\left| \frac{u_f}{u_o} \right| = e^{-\frac{2B}{m}x}\end{equation}

am I typing something wrong?

\left|\frac{u_f}{u_o}\right| = e^{-\frac{2B}{m}x}

\begin{equation} \ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x \end{equation}

\begin{equation} \ln|u_f| - \ln|u_o| = -\frac{2B}{m}x \end{equation}

so let me think about this. u is velocity from 54 m/s to 0 m/s. xf is what I'm solving for. Would I only integrate the LHS?

\begin{equation} \ln|u| = -\frac{2B}{m}x \end{equation}

?

but there is two answers. a + and a - since we were taking the square root

(1/2) is to the power. After I get u(x) then I change it back to v then do I integrate? 120 to 0 is bounds for velocity and I need to find ∆x which is just xf-xi which are the bounds for dx(position)