Question

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?

• A. Path of the particle is a circle of radius $4$ $m$
• B. Acceleration vector is along $-\overrightarrow{R}$
• C. Magnitude of acceleration vector is $\frac{v^2}{R}$, where $v$ is the velocity of particle
• D. Magnitude of the velocity of particle is $8$ $m{s}^{-2}$

Answer 1

The correct option is B Acceleration vector is along $-\overrightarrow{R}$

$\overrightarrow{R}=4\sin \left(2\pi t\right)\hat{i}+4\cos \left(2\pi t\right)\hat{j}Thisisthevectorequationv=\frac{dR}{dt}=8\pi \cos \left(2\pi t\right)\hat{i}-8\pi \sin \left(2\pi t\right)\hat{j}a=\frac{dV}{dt}=-16{\pi }^2\sin \left(2\pi t\right)\hat{i}-16{\pi }^2\cos \left(2\pi t\right)\hat{j}a=-4{\pi }^2\left(4\sin \right(2\pi t)\hat{i}+\cos (2\pi t)\widehat{j)}=-4{\pi }^2(\overrightarrow{R})Hence,theaccelerationvectorisalong-\overrightarrow{R}$