Question

The position co-ordinates of a particle moving in a $3D$ coordinate system is given by $x=a\cos \omega t,$ $y=a\sin \omega t$ and $z=a\omega t$. The speed of the particle is

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

Answer 1

The correct option is D $\sqrt{2}a\omega$

Speed of particle in $3D$ is given by
$v=\sqrt{v_x^2+v_y^2+v_z^2}.....\left(1\right)$

Here,
$v_x=\frac{dx}{dt}=\frac{d\left(a\cos \omega t\right)}{dt}=-a\omega \sin \omega t$
$v_y=\frac{dy}{dt}=\frac{d\left(a\sin \omega t\right)}{dt}=a\omega \cos \omega t$
and
$v_z=\frac{dz}{dt}=\frac{d\left(a\omega t\right)}{dt}=a\omega$
Using equation $1,$ we get

$\Rightarrow v=\sqrt{(-a\omega \sin \omega t{)}^2+(a\omega \cos \omega t{)}^2+(a\omega {)}^2}$
$\Rightarrow v=\sqrt{2}a\omega$