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The position vector of a particle $\vec R$ as a function of time is given by $\overrightarrow {\;R} = 4\sin \left( {2\pi t} \right)\hat i + 4\cos \left( {2\pi t} \right)\hat j$ where $R$ is in meters, $t$ is in seconds and $\hat i$ and $\hat j$ denote unit vectors along $x-$ and $y-$directions, respectively. Which one of the following statements is wrong for the motion of particle?
Path of the particle is a circle of radius $4$ meter.
Acceleration vector is along $-\overrightarrow {\;R} $
Magnitude of acceleration vector is $\frac{{{V^2}}}{R}\;$ where $ V$ is the velocity of particle.
Magnitude of the velocity of particle is $8 \ m/s $
Solution
$\begin{array}{l}
\,Here,\,\bar R = 4\sin \left( {2\pi t} \right)\hat i + 4\cos \left( {2\pi t} \right)\hat j\\
The\,velocity\,of\,the\,particle\,is\\
\bar v = \frac{{d\bar r}}{{dt}} = \frac{d}{{dt}}\left[ {4\sin \left( {2\pi t} \right)\hat i + 4\cos \left( {2\pi t} \right)\hat j} \right]\\
= \,8\pi \cos \left( {2\pi t} \right)\hat i – 8\pi \sin \left( {2\pi t} \right)\hat j\\
Its\,magnitude\,is
\end{array}$
$\begin{array}{l}
\left| {\bar v} \right| = \sqrt {{{\left( {8\pi \cos \left( {2\pi t} \right)} \right)}^2} + {{\left( { – 8\pi \sin \left( {2\pi t} \right)} \right)}^2}} \\
\,\,\,\,\,\, = \sqrt {64{\pi ^2}{{\cos }^2}\left( {2\pi t} \right)64{\pi ^2}{{\sin }^2}\left( {2\pi t} \right)} \\
\,\,\,\,\,\, = \sqrt {64{\pi ^2}\left[ {{{\cos }^2}\left( {2\pi t} \right) + {{\sin }^2}\left( {2\pi t} \right)} \right]} \\
\,\,\,\,\,\, = \sqrt {64{\pi ^2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {as\,{{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right)\\
\,\,\,\,\,\, = 8\pi \,m/s
\end{array}$
