Question

The position vector of a particle $\vec{R}$ as a function of time is given by $\vec{R}=4 \sin (2 \pi t) \hat{i}+4 \cos (2 \pi t) \hat{j}$ 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. Magnitude of the velocity of particle is 8 meter/second
• B. Path of the particle is a circle of radius 4 meter
• C. Acceleration vector is along $-\vec{R}$
• D. Magnitude of acceleration vector is $ \frac{ v^2}{ R}$
  where v is the velocity of particle.

Answer 1

Answer: (A)

Here, $ \vec{R} = 4 \sin ( 2 \pi t) \widehat{i} + 4 \cos ( 2 \pi t) \widehat{j}$
The velocity of the particle is
$ \vec{v} = \frac{ d \vec{R}}{ dt} = \frac{ d}{ dt} [ 4 \sin ( 2 \pi t ) \widehat{i} + 4 \cos ( 2 \pi t) \widehat{j} ] $
= 8 $ \pi \cos (2 \pi t) \widehat{i} - 8 \pi \sin ( 2 \pi t) \widehat{j} $
Its magnitude is
$ | \vec{v} | = \sqrt{ ( 8 \pi \cos ( 2 \pi t))^2 + ( - 8 \pi \sin (2 \pi t ))^2 } $
= $ \sqrt{ 64 \pi^2 \, \cos^2 (2 \pi t) + 64 \pi^2 \, \sin^2 (2 \pi t)}$
= $ \sqrt{ 64 \pi^2 \, [\cos^2 (2 \pi t) + \sin^2 (2 \pi t)]}$
= $ \sqrt{ 64 \pi^2}$ $ (as \, \sin^2 \theta + \cos^2 \theta = 1 ) $
$= 8 \pi \,m / s$