समय के फलन के रूप में किसी कण का स्थिति सदिश | vedclass

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Question

$\begin{array}{l}
\,Here,\,\bar R = 4\sin \left( {2\pi t} ight)\hat i + 4\cos \left( {2\pi t} ight)\hat j\
The\,velocity\,of\,the\,particle\,is

Vedclass · Accepted Answer

कण के वेग का परिमाण $8$ मी./से. है|

Vedclass · Answer

$\begin{array}{l}
\,Here,\,\bar R = 4\sin \left( {2\pi t} ight)\hat i + 4\cos \left( {2\pi t} ight)\hat j\
The\,velocity\,of\,the\,particle\,is\
\bar v = \frac{{d\bar r}}{{dt}} = \frac{d}{{dt}}\left[ {4\sin \left( {2\pi t} ight)\hat i + 4\cos \left( {2\pi t} ight)\hat j} ight]\
 = \,8\pi \cos \left( {2\pi t} ight)\hat i - 8\pi \sin \left( {2\pi t} ight)\hat j\
Its\,magnitude\,is
\end{array}$

$\begin{array}{l}
\left| {\bar v} ight| = \sqrt {{{\left( {8\pi \cos \left( {2\pi t} ight)} ight)}^2} + {{\left( { - 8\pi \sin \left( {2\pi t} ight)} ight)}^2}} \
\,\,\,\,\,\, = \sqrt {64{\pi ^2}{{\cos }^2}\left( {2\pi t} ight)64{\pi ^2}{{\sin }^2}\left( {2\pi t} ight)} \
\,\,\,\,\,\, = \sqrt {64{\pi ^2}\left[ {{{\cos }^2}\left( {2\pi t} ight) + {{\sin }^2}\left( {2\pi t} ight)} ight]} \
\,\,\,\,\,\, = \sqrt {64{\pi ^2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {as\,{{\sin }^2}	heta  + {{\cos }^2}	heta  = 1} ight)\
\,\,\,\,\,\, = 8\pi \,m/s
\end{array}$

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