Q. The position co-ordinates of a particle moving in a 3-D 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 :

Solution:

$v_{x} = - a \omega \sin\omega t \Rightarrow v_{y} = a \omega \cos\omega t $
$ v_{z} = a \omega \Rightarrow v = \sqrt{v^{2}_{x} + v^{2}_{y} +v^{2}_{z}} $
$ v = \sqrt{2} a \omega$

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions