The position vector of a particle is `vec( r) = a cos omega t i + a sin omega t j`, the velocity of the particle is

Options

A. parallel to the position vector
B. perpendicular to the position vector
C. directed towards the origin
D. directed away from the origin

Explanation

To find the velocity of the particle given its position vector, we can follow these steps:

### Step 1: Write down the position vector
The position vector of the particle is given as:
\[ \vec{r} = a \cos(\omega t) \hat{i} + a \sin(\omega t) \hat{j} \]

### Step 2: Differentiate the position vector with respect to time
To find the velocity vector \(\vec{v}\), we differentiate the position vector \(\vec{r}\) with respect to time \(t\):
\[ \vec{v} = \frac{d\vec{r}}{dt} \]

### Step 3: Differentiate each component
Now, we differentiate each component of the position vector:
- For the \(x\)-component:
\[ \frac{d}{dt}(a \cos(\omega t)) = -a \omega \sin(\omega t) \]
- For the \(y\)-component:
\[ \frac{d}{dt}(a \sin(\omega t)) = a \omega \cos(\omega t) \]

### Step 4: Combine the differentiated components
Thus, the velocity vector becomes:
\[ \vec{v} = -a \omega \sin(\omega t) \hat{i} + a \omega \cos(\omega t) \hat{j} \]

### Step 5: Factor out common terms
We can factor out \(a \omega\):
\[ \vec{v} = a \omega (-\sin(\omega t) \hat{i} + \cos(\omega t) \hat{j}) \]

### Step 6: Analyze the relationship between \(\vec{v}\) and \(\vec{r}\)
To determine the relationship between the velocity vector \(\vec{v}\) and the position vector \(\vec{r}\), we can compute the dot product:
\[ \vec{v} \cdot \vec{r} \]

### Step 7: Compute the dot product
The dot product is given by:
\[ \vec{v} \cdot \vec{r} = (a \omega (-\sin(\omega t)))(a \cos(\omega t)) + (a \omega \cos(\omega t))(a \sin(\omega t)) \]
This simplifies to:
\[ = -a^2 \omega \sin(\omega t) \cos(\omega t) + a^2 \omega \sin(\omega t) \cos(\omega t) = 0 \]

### Step 8: Conclusion about the relationship
Since the dot product \(\vec{v} \cdot \vec{r} = 0\), this indicates that the velocity vector \(\vec{v}\) is perpendicular to the position vector \(\vec{r}\).

### Final Answer
The velocity of the particle is:
\[ \vec{v} = a \omega (-\sin(\omega t) \hat{i} + \cos(\omega t) \hat{j}) \]
And it is perpendicular to the position vector \(\vec{r}\).

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