A particle moves so that its position vector is given by `vec r = cos omega t hat x + sin omega t hat y`, where `omega` is a constant which of the following is true ?

Options

A. Velocity and acceleration both are parallel to r
B. velocity is perpendicular to r and acceleration is directed towards to origin
C. Velocity is perpendicular to r and acceleration is directed away from the origin
D. Velocity and acceleration both are perpendicular to r

Explanation

To solve the problem, we need to analyze the position vector of the particle given by:

\[ \vec{r} = \cos(\omega t) \hat{x} + \sin(\omega t) \hat{y} \]

where \(\omega\) is a constant. We will find the velocity and acceleration vectors and then check the statements provided in the options.

### Step 1: Find the Velocity Vector

The velocity vector \(\vec{v}\) is the time derivative of the position vector \(\vec{r}\):

\[ \vec{v} = \frac{d\vec{r}}{dt} \]

Differentiating \(\vec{r}\):

\[ \vec{v} = \frac{d}{dt}(\cos(\omega t) \hat{x} + \sin(\omega t) \hat{y}) \]

Using the chain rule:

\[ \vec{v} = -\sin(\omega t) \cdot \omega \hat{x} + \cos(\omega t) \cdot \omega \hat{y} \]

This can be written as:

\[ \vec{v} = -\omega \sin(\omega t) \hat{x} + \omega \cos(\omega t) \hat{y} \]

### Step 2: Find the Acceleration Vector

The acceleration vector \(\vec{a}\) is the time derivative of the velocity vector \(\vec{v}\):

\[ \vec{a} = \frac{d\vec{v}}{dt} \]

Differentiating \(\vec{v}\):

\[ \vec{a} = \frac{d}{dt}(-\omega \sin(\omega t) \hat{x} + \omega \cos(\omega t) \hat{y}) \]

Using the chain rule again:

\[ \vec{a} = -\omega \cos(\omega t) \cdot \omega \hat{x} - \sin(\omega t) \cdot \omega^2 \hat{y} \]

This simplifies to:

\[ \vec{a} = -\omega^2 \cos(\omega t) \hat{x} - \omega^2 \sin(\omega t) \hat{y} \]

We can factor out \(-\omega^2\):

\[ \vec{a} = -\omega^2 (\cos(\omega t) \hat{x} + \sin(\omega t) \hat{y}) = -\omega^2 \vec{r} \]

### Step 3: Analyze the Relationships

1. **Velocity and Position Vector**: To check if \(\vec{v}\) and \(\vec{r}\) are perpendicular, we compute the dot product \(\vec{v} \cdot \vec{r}\):

\[ \vec{v} \cdot \vec{r} = (-\omega \sin(\omega t) \hat{x} + \omega \cos(\omega t) \hat{y}) \cdot (\cos(\omega t) \hat{x} + \sin(\omega t) \hat{y}) \]

Calculating the dot product:

\[ = -\omega \sin(\omega t) \cos(\omega t) + \omega \cos(\omega t) \sin(\omega t) = 0 \]

Since \(\vec{v} \cdot \vec{r} = 0\), \(\vec{v}\) is perpendicular to \(\vec{r}\).

2. **Acceleration and Position Vector**: To check if \(\vec{a}\) and \(\vec{r}\) are perpendicular, we compute the dot product \(\vec{a} \cdot \vec{r}\):

\[ \vec{a} \cdot \vec{r} = (-\omega^2 (\cos(\omega t) \hat{x} + \sin(\omega t) \hat{y})) \cdot (\cos(\omega t) \hat{x} + \sin(\omega t) \hat{y}) \]

Calculating the dot product:

\[ = -\omega^2 (\cos^2(\omega t) + \sin^2(\omega t)) = -\omega^2 \]

Since \(\vec{a} \cdot \vec{r} \neq 0\), \(\vec{a}\) is not perpendicular to \(\vec{r}\).

### Conclusion

From the analysis, we conclude:

- The velocity \(\vec{v}\) is perpendicular to the position vector \(\vec{r}\).
- The acceleration \(\vec{a}\) is directed towards the origin (as it is proportional to \(-\vec{r}\)).

Thus, the correct statement is that the velocity is perpendicular to the position vector and the acceleration is directed towards the origin.

### Final Answer

The correct option is **Option 2**: "Velocity is perpendicular to r and acceleration is directed towards the origin."