Question

In three dimensional system, the position coordinates of a particle (in motion) are given below
$x =a \cos \omega t$
$y =a \sin \omega t$
$z =a \omega t$
The velocity of particle will be

• A. $\sqrt{2} a \omega$
• B. $2 a \omega$
• C. $a \omega$
• D. $\sqrt{3} a \omega$

Answer 1

Answer: (A)

Given that the position coordinates of a particle

So, the position vector of the particle is
$\hat{ r } =x \hat{ i }+y \hat{ j }+z \hat{ k }$
$\Rightarrow \hat{ r } =a \cos \omega t \hat{ i }+a \sin \omega t \hat{ j }+a \omega t \hat{ k } $
$\hat{ r } =a[\cos \omega t \hat{ i }+\sin \omega t \hat{ j }+\omega t \hat{ k }] $
therefore, the velocity of the particle is
$\because \hat{ v }=\frac{d r }{d t}=\frac{d[a][\cos \omega t \hat{ i }+\sin \omega t \hat{ j }+\omega t \hat{ k }]}{d t} $
$\Rightarrow \hat{ v }=-a \omega \sin \omega t \hat{ i }+a \omega \cos \omega t \hat{ j }+a \omega \hat{ k }) $
The magnitude of velocity is
$| v |=\sqrt{v_{x}^{2}+v_{y}^{2}+v_{z}^{2}} $
or $ | v |=\sqrt{(-a \omega \sin \omega t)^{2}+(a \omega \cos \omega t)^{2}+(a \omega)^{2}}$
$=\omega a \sqrt{(-\sin \omega t)^{2}+(\cos \omega t)^{2}+(1)^{2}} $
$=\sqrt{2} \omega a$