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

Explanation

To solve the problem, we need to analyze the motion of a particle given its position vector:

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

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

### Step 1: Find the Velocity Vector

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

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

Calculating the derivative:

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

So, the velocity vector is:

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

### Step 2: Find the Acceleration Vector

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

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

Calculating the derivative:

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

So, the acceleration vector is:

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

### Step 3: Check if Velocity is Perpendicular to Position Vector

To check if \(\vec{v}\) is perpendicular to \(\vec{r}\), 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 the dot product is zero, \(\vec{v}\) is perpendicular to \(\vec{r}\).

### Step 4: Check if Acceleration is Directed Toward the Origin

We can express the acceleration vector in terms of the position vector \(\vec{r}\):

\[ \vec{a} = -\omega^2 \vec{r} \]

This indicates that the acceleration vector is directed towards the origin (since it is in the opposite direction to \(\vec{r}\)).

### Conclusion

Based on our calculations:

1. The velocity vector \(\vec{v}\) is perpendicular to the position vector \(\vec{r}\).
2. The acceleration vector \(\vec{a}\) is directed toward the origin.

Thus, the correct statement is that the velocity is perpendicular to \(\vec{r}\) and the acceleration is directed toward the origin.