The position vector of the particle is `r(t)=acos omegat hati+a sin omega thatj`, where a and `omega` are real constants of suitable dimensions. The acceleration is

Options

A. Perpendicular to the velocity
B. Parallel to the velocity
C. Directed away from the origin
D. Perpendicular to the position vector

Answer

To find the acceleration of the particle given its position vector \( \mathbf{r}(t) = a \cos(\omega t) \hat{i} + a \sin(\omega t) \hat{j} \), we will follow these steps:

### Step 1: Differentiate the Position Vector to Find Velocity
The first step is to differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \) to find the velocity vector \( \mathbf{v}(t) \).

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

Using the chain rule, we differentiate each component:

- For the \( \hat{i} \) component:
\[ \frac{d}{dt}(a \cos(\omega t)) = -a \omega \sin(\omega t) \]

- For the \( \hat{j} \) component:
\[ \frac{d}{dt}(a \sin(\omega t)) = a \omega \cos(\omega t) \]

So, the velocity vector is:
\[ \mathbf{v}(t) = -a \omega \sin(\omega t) \hat{i} + a \omega \cos(\omega t) \hat{j} \]

### Step 2: Differentiate the Velocity Vector to Find Acceleration
Next, we differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \) to find the acceleration vector \( \mathbf{a}(t) \).

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

Again, using the chain rule, we differentiate each component:

- For the \( \hat{i} \) component:
\[ \frac{d}{dt}(-a \omega \sin(\omega t)) = -a \omega^2 \cos(\omega t) \]

- For the \( \hat{j} \) component:
\[ \frac{d}{dt}(a \omega \cos(\omega t)) = -a \omega^2 \sin(\omega t) \]

So, the acceleration vector is:
\[ \mathbf{a}(t) = -a \omega^2 \cos(\omega t) \hat{i} - a \omega^2 \sin(\omega t) \hat{j} \]

### Step 3: Express Acceleration in Terms of Position Vector
Notice that the acceleration vector can be expressed in terms of the position vector \( \mathbf{r}(t) \):
\[ \mathbf{a}(t) = -\omega^2 (a \cos(\omega t) \hat{i} + a \sin(\omega t) \hat{j}) = -\omega^2 \mathbf{r}(t) \]

### Conclusion
Thus, the acceleration of the particle is given by:
\[ \mathbf{a}(t) = -\omega^2 \mathbf{r}(t) \]