Position of particle as a function of time is given as `vec r=cos wt hati+sin wt hatj` . Choose correct statement about `vecr,vec v` and `vec a` where `vec v` and `vec a` are velocity and acceleration of particle at time t.

Options

A. `vec v ` is perpendicular to `vec r` and `vec a` is towards origin
B. `vec v` and `vec a` are perpendicular to `vec r`
C. `vec v` is parallel to `vec r` and `vec a` parallel to `vec r`
D. `vec v` is perpendicular to `vec r` and `vec a` is away from origin

Explanation

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

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

### Step 1: Find the velocity vector \(\vec{v}\)

The velocity vector \(\vec{v}\) can be found by differentiating the position vector \(\vec{r}\) with respect to time \(t\):

\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}) \]

Using the chain rule, we differentiate each component:

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

Thus, we can express the velocity vector as:

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

### Step 2: Find the acceleration vector \(\vec{a}\)

The acceleration vector \(\vec{a}\) can be found by differentiating the velocity vector \(\vec{v}\) with respect to time \(t\):

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

Again, applying the chain rule:

\[ \vec{a} = -\omega \cos(\omega t) \cdot \omega \hat{i} - \sin(\omega t) \cdot \omega \hat{j} \]

This simplifies to:

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

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

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

### Step 3: Analyze the relationships between \(\vec{r}\), \(\vec{v}\), and \(\vec{a}\)

1. **Check if \(\vec{v}\) is perpendicular to \(\vec{r}\)**:
To check if \(\vec{v}\) is perpendicular to \(\vec{r}\), we calculate the dot product \(\vec{v} \cdot \vec{r}\):

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

Expanding this gives:

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

Since the dot product is zero, \(\vec{v}\) is indeed perpendicular to \(\vec{r}\).

2. **Check if \(\vec{a}\) is antiparallel to \(\vec{r}\)**:
We have found that:

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

This indicates that \(\vec{a}\) is in the opposite direction to \(\vec{r}\), confirming that \(\vec{a}\) is antiparallel to \(\vec{r}\).

### Conclusion

From our analysis, we conclude that:

- \(\vec{v}\) is perpendicular to \(\vec{r}\).
- \(\vec{a}\) is antiparallel to \(\vec{r}\).

Thus, the correct statement about \(\vec{r}\), \(\vec{v}\), and \(\vec{a}\) is that \(\vec{v}\) is perpendicular to \(\vec{r}\) and \(\vec{a}\) is directed towards the origin (antiparallel to \(\vec{r}\)).