The position vector of a particle $\overrightarrow{R}$ 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?

The correct option is(D)

Given,

Position vector $\overrightarrow{R}=4sin\left(2\pi t\right)\overrightarrow{i}+4cos\left(2\pi t\right)\overrightarrow{j}$

For finding the velocity vector we have to differentiate it with respect to the time

So,

$\frac{dR}{dt}=\overrightarrow{v}=8\pi cos\left(2\pi t\right)\overrightarrow{i}-8\pi sin\left(2\pi t\right)\overrightarrow{j}$

For finding the acceleration we have to differentiate the velocity with respect to the time

So,

$\frac{dv}{dt}=\overrightarrow{a}=-4{\left(2\pi \right)}^{2}sin\left(2\pi \right)\overrightarrow{i}-4{\left(2\pi \right)}^{2}cos\left(2\pi t\right)\overrightarrow{j}=-{\left(2\pi \right)}^2\overrightarrow{R}$

So we can say that the acceleration is in the direction of $\overrightarrow{-R}$

Now we have to find the magnitude of the velocity

Magnitude of the velocity = $\sqrt{{\left(8\pi cos\left(2\pi t\right)\right)}^2-{\left(8\pi sin\left(2\pi t\right)\right)}^2}$

After solving this we get

The magnitude of the velocity = $8\pi$

Now,

Magnitude of the acceleration = $|-{\left(2\pi \right)}^2\overrightarrow{R}|=16{\pi }^2=\frac{{\left(8\pi \right)}^2}{4}=\frac{v^2}{R}$

Hence from all the above findings, we can say that the only option (D) is wrong