The position vectors os a particle is `r=(acosomegat)hati+(asinomegat)hatj.`The velocity of particle is

Options

A. directiont towards the origin
B. directionaway from the origin
C. parallelto the position vector
D. perpendicular to the position vector

Answer

To solve the problem, we need to determine the velocity of the particle given its position vector. The position vector is given as:

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

### Step 1: Differentiate the Position Vector
To find the velocity vector, we differentiate the position vector \(\mathbf{r}\) with respect to time \(t\):

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

### Step 2: Apply the Derivative
Now, we differentiate each component of \(\mathbf{r}\):

1. Differentiate \(a \cos(\omega t)\):
\[ \frac{d}{dt}(a \cos(\omega t)) = -a \omega \sin(\omega t) \]

2. Differentiate \(a \sin(\omega t)\):
\[ \frac{d}{dt}(a \sin(\omega t)) = a \omega \cos(\omega t) \]

Putting these together, we get the velocity vector:

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

### Step 3: Analyze the Relationship Between Velocity and Position
Next, we need to analyze the relationship between the velocity vector \(\mathbf{v}\) and the position vector \(\mathbf{r}\).

1. The position vector is:
\[ \mathbf{r} = a \cos(\omega t) \hat{i} + a \sin(\omega t) \hat{j} \]

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

### Step 4: Calculate the Dot Product
To determine the relationship between the velocity and position vectors, we calculate the dot product \(\mathbf{v} \cdot \mathbf{r}\):

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

Calculating this gives:

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

\[ = -a^2 \omega \sin(\omega t) \cos(\omega t) + a^2 \omega \sin(\omega t) \cos(\omega t) = 0 \]

### Step 5: Conclusion
Since the dot product \(\mathbf{v} \cdot \mathbf{r} = 0\), this indicates that the velocity vector is perpendicular to the position vector.

Thus, the final answer is that the velocity of the particle is **perpendicular to the position vector**.

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