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\begin{document}

\title{Soluci\selectlanguage{ngerman}ón de ejercicios \#2\selectlanguage{english}}


\author{esau salas }
\affil{Instituto Tecnológico Superior Zacatecas Occidente}



\date{\today}

\maketitle

\selectlanguage{english}
\begin{abstract}
solucion de ejercicios de la unidad 2%
\end{abstract}%



\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Captura/Captura1}
\caption{{(a)
{\label{958713}}%
}}
\end{center}
\end{figure}

(a)~

sen\(\theta\)=3/5

tan\(\theta=\)3/4

\(\theta\)=36.87\selectlanguage{ngerman}°

\(\cos\theta=\frac{4}{5}\)\(\Sigma fy=\)\(N-500\ \sin\theta-w=0\)

\(N=500\left(\frac{3}{6}\right)N=98.1N\)

\(N=398.1N\)

se sacaron los 3 angulos seno,coseno y tangente y los 3 da el mismo
resultado y se ase la sumatoria de fuerzas en (x) y (y)

\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Captura1/Captura2}
\caption{{(b)
{\label{148158}}%
}}
\end{center}
\end{figure}

\par\null\par\null

(b)

\(\Sigma Fx=\)

\(500\cos\theta N-300N=m\alpha\)

\(400N-300N=m\alpha\)

\(\frac{10m}{s^2}=\alpha\)

\(\alpha=\frac{da}{dt}\)

\(\alpha=n.+\alpha t\)

\(\alpha=\alpha t\)

\(\alpha=\frac{20m}{s^{ }}\)

Una particula viaja a lo largo de una linea recta con una velocidad de
v=(4(-3t2)m/s donde t esta en segundos.determine la posicion de la
particula cuando t=4s considere x=0 cuando t=0

\par\null

\(x=2t^2-t^3\)

t=4s

a=4-6t

\(x=2\left(4\right)^2-\left(4\right)^3\)

\(x=32-64\)

x=-32m

\(v=\frac{d}{t}\)

d=vt

\(v=\int_{at-3t^2}^{ }\)

\(4\int_{t\ dt-3\int_{t^{2\ \ dt}}^t}^t\)

\(4\left(\frac{t^2}{2}\right)-3\left(\frac{t^3}{3}\right)\)

\(x=2t^2-t^3\)

\par\null

\par\null

\selectlanguage{english}
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