\documentclass{article}
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\fancyhead[LO,LE]{Carter Schultz, Dynamics, HW3}
\fancyfoot[C]{\bf \copyright\  2012 C. J. Schultz}
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\begin{document}
\section{Problem 12-74}
\textbf{Given:} The velocity of a particle is given by $v = (16t^2 i + 4t^3j +(5t+2)k)$ m/s where, $t$ is in seconds. If the particle is at the origin when $t=0$, determine the magnitude of the particle's acceleration when $t=2$s. Also, what is the $x,y,z$ coordinate position of the particle at this instant.
\[\frac{d\vec{v}}{dt} = \vec{a}\]
\[ \vec{a} =(32t\;i +12t^2\;j + 5\;k)   \]
\[\frac{d\vec{s}}{dt} = \vec{v}\]
\[d\vec{s} = \vec{v}\;dt\]
\[\int{d\vec{s}} = \int{(16t^2 i + 4t^3j +(5t+2)k)}dt\]
\[\vec{s} = \frac{16}{3} t^3i +t^4j+(\frac{5}{2}t^2 + 2t)k + C\]
\[\vec{a}(2) = (64i+48j+5k)\]
\[\norm{\vec{a}(2)} = 5\sqrt{257}\]
\[\norm{\vec{a}(2)} = 80.2 [m/s^2]\]
\[\vec{s}(2)=(128/3i +16j +14k)\]
\textcolor{red}{Note: Defaulting to 3 sig figs.}

\newpage

\section{Problem 12-78}
\textbf{Given:} Pegs A and B are restricted to move in the elliptical slots due to the motion of the slotted link. If the link moves with a constant speed of 10m/s, determine the magnitude of the velocity and acceleration of peg A when x = 1m.
\begin{center}
  \includegraphics[height=2in]{Pic1.png}
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\[ \frac{dx}{dt} = 10 [m/s]\]
\[\frac{d}{dt}\left(\frac{x^2}{4} + y^2\right) = \frac{d}{dt}(1) \]
\[\frac{1}{2}x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\]
solving for $\frac{dy}{dt}$
\[\frac{dy}{dt}=\frac{x}{4y}\frac{dx}{dt}\]
\[\frac{dy}{dt} =-\frac{10x}{4y}\]
\[\vec{v} = \left\{ \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j}  \right\}\]
\[ \vec{v} = \left\{ 10 \hat{i} - \frac{10x}{4y}\hat{j} \right\} \]
Solving for $\vec{a}$
\[ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d(10)}{dt} \hat{i} - \frac{d}{dt}\left(\frac{10x}{4y}\right) \hat{j}\]
applying the chain rule and substituting previous solutions back in yields:
\[\vec{a} = \frac{d\vec{v}}{dt} = 0 \hat{i} - \left(\frac{25}{y} - \frac{25x^2}{4y^3}\right) \hat{j}\]
when x equals 1:
\[y=\sqrt{.75}=\sqrt{1- \frac{1^2}{4}}\]
\[\frac{dy}{dt} = \frac{10}{4\sqrt{.75}}\]
\[\norm{\vec{v}(1,4)} = \norm{10i - \frac{10}{4\sqrt{.75}}j} =\fbox{10.4 [m/s]}\] 
\[\norm{\frac{d\vec{v}}{dt}} = \norm{ 0 \hat{i} - \left(\frac{25}{\sqrt{.75}} - \frac{25(1)^2}{4\sqrt(.75)^3}\right) \hat{j}} = \fbox{38.5 [m/(s*s)]} \]
\textcolor{red}{Note: Defaulting to 3 sig figs.}


\newpage

\section{Problem 12-86}
\textbf{Given:} The motorcycle travels with a constant speed $v_0$ along the path that, for a short distance, takes the form of a sine curve. Determine the $x$ and $y$ components of its velocity at any instant on the curve.
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  \includegraphics[height=2in]{Pic2.png}
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\begin{equation}
v_0 = \sqrt{\frac{dx}{dt}^2 +\frac{dx}{dt}^2}  
\end{equation}
\begin{equation}
\frac{dy}{dt} = v_y = \frac{d}{dt} \left( c \sin\left(\frac{\pi}{L} x\right) \right) = c\left(\cos{\left(\frac{\pi}{L}x\right)}\frac{\pi}{L}\frac{dx}{dt}\right)   
\end{equation}
substituting this into the origional equation 1 yields:
\begin{equation}
v_0 = \sqrt{\frac{dx}{dt}^2 \left(1+\left(\cos{\left(\frac{\pi}{L}x\right)}\frac{\pi c}{L} \right)^2\right)}
\end{equation}
solving this for $\frac{dx}{dt}$ yields:
\begin{equation}
\frac{dx}{dt} = v_0 \sqrt{1+\left(\cos{\left(\frac{\pi}{L}x\right)}\frac{\pi c}{L} \right)^2}
\end{equation}
substituting equation 4 into equation 2 yields:
\begin{equation}
\frac{dy}{dt} = c\left(\cos{\left(\frac{\pi}{L}x\right)}\frac{\pi}{L}\right)v_0    \sqrt{1+\left(\cos{\left(\frac{\pi}{L}x\right)}\frac{\pi c}{L} \right)^2}
\end{equation}
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