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 D

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

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

So,

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

Hence we can say that the acceleration is along the -$\overrightarrow{R}$

If we find the magnitude of the velocity then we will get

$\left|\overrightarrow{v}\right|=8\pi$

If we find the magnitude of the acceleration vector then we will get

$\left|\overrightarrow{a}\right|=\frac{v^2}{R}$

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