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