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\fancyhead[LO,LE]{Carter Schultz, Dynamics, HW2}
\fancyfoot[C]{\bf \copyright\  2012 C. J. Schultz}
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\begin{document}
\section{Problem 12-24}
\textbf{Given:} A particle starts from rest and travels along a straight 
line with an acceleration $a = (30 - .2v)ft/s^2$, where $v$ is in ft/s.

\noindent \textbf{Determine:} The time when the velocity 
of the particle is $v=30ft/s$


\noindent \textbf{Solution:} 


\[a = (30 -.2v)[ft/s^2]\}\]
\[\frac{dv}{dt} = (30 -.2v)\]
\[\frac{dv}{(30-.2v)} =dt \]
\[\int\limits_{v_0}^v\frac{dv}{(30-.2v)} =\int dt \]
\[-5\ln{\abs{30-.2v}}=t+C\]
setting initial condition
\[-5\ln{\abs{30}}=C \]
\[-5\ln{\abs{30-.2v}}+5\ln{\abs{30}}=t\]
setting target condition
\[t=-5\ln{\abs{30-.2(30)}}+5\ln{\abs{30}}\]
\[\mathbf{\fbox{t =1.16 [s] }\;\;\overbrace{1.11572 [s]}} \; \text{\textcolor{red}{assuming 3 sig figs}}\]

\newpage

\section{Problem 12-36}
\textbf{Given:} The acceleration of a particle traveling along a striaght line is $a = (8-2s) [m/s^2]$, where $s$ is in meters. If $v=0$ at $s=0$, determine the velocity of the particle at $s=2[m]$, and the position of the particle when the velocity is maximum.


\noindent \textbf{Solution:}
\[a =\frac{dv}{ds}\frac{ds}{dt}\]
\[a\;ds =v\;dv\]
\[\int\limits_{s_0}^s(8-2s)ds=\int\limits_{v_0}^vv\;dv\]
\[s_0 = 0\; \&\; v_0 =0\]
\[8s-s^2 =\frac{v^2}{2}\]
\[v = \pm\frac{1}{2}\sqrt{8s-s^2}\]
\[v(2)=.5\sqrt{16-4}\]
\[\fbox{v(2)=1.73[m/s]}\; \text{\textcolor{red}{assuming 3 sig figs}}\]
\[\frac{dv}{dt}= a = 0 \;\text{to maximize}\]
\[8-2s=0\]
\[\fbox{s=4.00 \;\text{when v is maximized}}\]

\newpage

\section{Problem 12-42}
\textbf{Given:} The boat travels in a straight line with the acceleration described by $a-s$ graph. If it starts from rest construct the $v-s$ graph and determine the boat's maximum speed. What distance $s'$ does it travel before it stops?

\begin{center}
\includegraphics[height=1.7in]{Boat.png}  
\end{center}
\[a=
\begin{cases}
  -.02s+6 & : x\leq150 \\
  -4& : x> 150
\end{cases}
\]
\[\int\limits_{s_0}^s{
\begin{cases}
  -.02s+6 & : x\leq150 \\
  -4& : x> 150
\end{cases}}ds = \int\limits_{v_0}^{v}{v\;dv}
\]
\[ v_0=0 \;\text{and}\; s_0 =0\]
\[
\begin{cases}
  -.01s^2+6s & : x\leq150 \\
  -4(s-150) + \frac{v(150)^2}{2} & : x> 150
\end{cases} = \frac{v^2}{2}
\]
\[v(150) = -.01(150)^2 + 6(150) = \fbox{675 [m/s]} \leftarrow \; \text{v max}\]
\[v = 
\begin{cases}
 \sqrt{ -.02s^2+12s} & : x\leq150 \\
  \sqrt{-8(s-150) + 1350}  & : x> 150
\end{cases}
\]
for v = 0 finding maximum ``s'' of second verison of the equation yeilds
\[\fbox{s = 318.75 [m]}\]


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