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De snelheid \PYG{l+s}{\PYGZdl{}}\PYG{n+nb}{ v }\PYG{l+s}{\PYGZdl{}} is gedefinieerd als
\PYG{k}{\PYGZbs{}begin}\PYG{n+nb}{\PYGZob{}}align\PYG{n+nb}{\PYGZcb{}}
    v \PYG{n+nb}{\PYGZam{}}:= \PYG{k}{\PYGZbs{}dod}\PYG{n+nb}{\PYGZob{}}x\PYG{n+nb}{\PYGZcb{}}\PYG{n+nb}{\PYGZob{}}t\PYG{n+nb}{\PYGZcb{}}
\PYG{k}{\PYGZbs{}end}\PYG{n+nb}{\PYGZob{}}align\PYG{n+nb}{\PYGZcb{}}
De oplossing van de differentiaalvergelijking
\PYG{l+s}{\PYGZdl{}}\PYG{n+nb}{ }\PYG{n+nv}{\PYGZbs{}frac}\PYG{n+nb}{\PYGZob{}}\PYG{n+nv}{\PYGZbs{}dif}\PYG{n+nb}{ v}\PYG{n+nb}{\PYGZcb{}}\PYG{n+nb}{\PYGZob{}}\PYG{n+nv}{\PYGZbs{}dif}\PYG{n+nb}{ t}\PYG{n+nb}{\PYGZcb{}}\PYG{n+nb}{ }\PYG{o}{=}\PYG{n+nb}{ }\PYG{n+nv}{\PYGZbs{}cos}\PYG{n+nb}{\PYGZca{}}\PYG{l+m}{2}\PYG{o}{(}\PYG{n+nb}{t}\PYG{o}{)}\PYG{n+nb}{ }\PYG{l+s}{\PYGZdl{}} is
\PYG{k}{\PYGZbs{}begin}\PYG{n+nb}{\PYGZob{}}align\PYG{n+nb}{\PYGZcb{}}
    v(t) \PYG{n+nb}{\PYGZam{}}= v\PYG{n+nb}{\PYGZus{}}0 + \PYG{k}{\PYGZbs{}int}\PYG{n+nb}{\PYGZus{}}\PYG{n+nb}{\PYGZob{}}0\PYG{n+nb}{\PYGZcb{}}\PYG{n+nb}{\PYGZca{}}\PYG{n+nb}{\PYGZob{}}t\PYG{n+nb}{\PYGZcb{}}\PYG{k}{\PYGZbs{}cos}\PYG{n+nb}{\PYGZca{}}2(t)\PYG{k}{\PYGZbs{}dif} t.
    \PYG{k}{\PYGZbs{}label}\PYG{n+nb}{\PYGZob{}}eq:exprVelocity\PYG{n+nb}{\PYGZcb{}}
\PYG{k}{\PYGZbs{}end}\PYG{n+nb}{\PYGZob{}}align\PYG{n+nb}{\PYGZcb{}}

...

Hiermee vinden we (\PYG{k}{\PYGZbs{}ref}\PYG{n+nb}{\PYGZob{}}eq:exprVelocity\PYG{n+nb}{\PYGZcb{}}) als
\PYG{k}{\PYGZbs{}begin}\PYG{n+nb}{\PYGZob{}}align*\PYG{n+nb}{\PYGZcb{}}
    v(t) \PYG{n+nb}{\PYGZam{}}= v\PYG{n+nb}{\PYGZus{}}0 + \PYG{k}{\PYGZbs{}frac}\PYG{n+nb}{\PYGZob{}}t\PYG{n+nb}{\PYGZcb{}}\PYG{n+nb}{\PYGZob{}}2\PYG{n+nb}{\PYGZcb{}}
    + \PYG{k}{\PYGZbs{}frac}\PYG{n+nb}{\PYGZob{}}1\PYG{n+nb}{\PYGZcb{}}\PYG{n+nb}{\PYGZob{}}4\PYG{n+nb}{\PYGZcb{}}\PYG{k}{\PYGZbs{}sin}(2t).
\PYG{k}{\PYGZbs{}end}\PYG{n+nb}{\PYGZob{}}align*\PYG{n+nb}{\PYGZcb{}}
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