1. Tardigrade
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  4. The position vector of a particle is vec r =(a cos ω t) hat i +(a sin ω t) hat j The velocity vector of the particle is:

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Q. The position vector of a particle is $\vec{ r }=(a \cos \omega t) \hat{ i }+(a \sin \omega t) \hat{ j }$ The velocity vector of the particle is:

Bihar CECEBihar CECE 2004Motion in a Straight Line

Solution:

Velocity is the rate of change of position vector,
$\vec{r}=a(\cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}$
Velocity vector, $\vec{v}=\frac{d \vec{r}}{d t}$
$\left.=\frac{d}{d t}[(a \cos \omega t)] \hat{i}+(a \sin \omega t) \hat{j}\right]$
$=(-a \sin \omega t) \hat{i}+(a \cos \omega t) \hat{j}$
Now, $\vec{v} \cdot \vec{r}=[(-a \sin \omega t) \hat{i}+(a \cos \omega t) \hat{j}][(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}]$
$=(-a \sin \omega t)(a \cos \omega t)+(a \cos \omega t)(a \sin \omega t)$
$=-a^{2} \sin \omega t \cos \omega t+a^{2} \cos \omega t \sin \omega t$
$=0$
As dot product of velocity vector with position vector is zero, so velocity vector of the particle is perpendicular to position vector.