Question

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

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

Answer 1

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

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

Given, position co-ordinates of a particle,

$x=a\cos \omega t$, $y=a\sin \omega t$ and $z=a\omega t$

So, position vector of the particle,

$\overrightarrow{r}=a\cos \omega t\hat{i}+a\sin \omega t\hat{j}+a\omega t\hat{k}$

So, velocity of the particle, $\overrightarrow{v}=\frac{d\overrightarrow{v}}{dt}$

$\overrightarrow{v}=\frac{d}{dt}(a\cos \omega t\hat{i}+a\sin \omega t\hat{j}+a\omega t\hat{k})$

$\overrightarrow{v}=-a\omega \sin \omega t\hat{i}+a\omega \cos \omega t\hat{i}+a\omega \hat{k}+(a\omega {)}^2$

So, speed of the particle.

$\left|\overrightarrow{v}\right|=\sqrt{(-a\omega \sin \omega t{)}^2+(a\omega \cos \omega t{)}^2+(a\omega {)}^2}$

$\left|\overrightarrow{v}\right|=\sqrt{2}a\omega$

Final answer :(c)