The position vector of a particle $\vec{R}$ as a funtion of time is given by:
$\vec{R}=4\sin \left(2\pi t\right)\hat{i}+4\cos \left(2\pi t\right)\hat{j}$
Where $R$ is in meters, $t$ is in seconds and $\hat{i}$ and $\hat{j}$ denote until vectors along x-and y- directions, respectively Which one of the following statements is wrong for the motion of particle ?

The position vector of a particle R as a
function of time is given by
$R=4\sin \left(2\pi t\right)\hat{i}+4\cos \left(2\pi t\right)\hat{j}$
x- axis component, x = 4 $\sin 2\pi t$
y-axis component,y = 4 $\cos 2\pi t$
Squaring and adding both equations we get
$x^2+y^2=4^{20({\sin }^2\left(2\pi t\right)+{\cos }^2\left(2\pi t\right)}$
i.e $x^2+y^2=4^2$i.e equations of circle and radius is 4 m
Acceleration vectors, $a=(\frac{v^2}{R}(-\hat{R}0.$ while v is velocity of a particle
Magnitude of acceleration vector $a=\frac{v^2}{R}$
As, we have $v_x=+4\left(\cos 2\pi t\right)2\pi$and
$v_y=-4\left(\sin 2\pi t\right)2\pi$
Net resultant velocity
$v=\sqrt{v_x^2+v_y^2}$
$=\sqrt{{\left(8\pi \right)}^2\left({\cos }^{2}2\pi t+{\sin }^{2}2\pi t\right)}$
$v=8\pi$
$\left|\therefore {\cos }^{2}2\pi t+{\sin }^{2}2\pi t=1\right|$
So, option (c0 is incorrect