Question

the position co-ordinates of a particle moving in a 3-D coordinate system is given
by $x=acos\omega t$
$x=asin\omega t$
and $z=a\omega t$
The speed of the particle is :

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

Answer 1

Answer: (B)

$\sqrt{2}a\omega$

The speed in the $x$ direction will be $V_x=\frac{d}{dt}x=-a\omega sin\omega t$

The speed in the $y$ direction will be $V_y=\frac{d}{dt}y=a\omega Cos\omega t$

The speed in the $z$ direction will be $V_z=\frac{d}{dt}z=a\omega$

So the net speed will be $magnitude$ of the $vector$ sum of the various components.

So $V=\sqrt{V_x^2+V_y^2+V_z^2}=\sqrt{a^2{\omega }^2(si{n}^2\omega t+co{s}^2\omega t)+a^2{\omega }^2}=a\omega \sqrt{2}$