A particle moves such that its momentum vector `vecp(t) = cos omega t hati + sin omega t hatj` where `omega` is a constant and t is time. Then which of the following statements is true for the velocity `vec v(t)` and acceleration `veca (t)` of the particle :

Options

A. `vecV` is perpendicular to `vecP` and `veca` is directed towards the origin
B. `vec v` and `veca` both are parallel to `vecp`
C. `vecv` and `veca` both are perpendicular to `vecp`
D. `vecv` is parallel to `vecp` and `veca` is perpendicular to `vecp`

Explanation

To solve the problem, we need to analyze the momentum vector given and derive the velocity and acceleration vectors from it.

### Step-by-Step Solution:

1. **Given Momentum Vector**:
The momentum vector is given as:
\[ \vec{p}(t) = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j} \]

2. **Relationship Between Momentum and Velocity**:
The momentum \(\vec{p}\) is related to velocity \(\vec{v}\) by the equation:
\[ \vec{p} = m \vec{v} \]
where \(m\) is the mass of the particle. Thus, the velocity vector can be expressed as:
\[ \vec{v}(t) = \frac{\vec{p}(t)}{m} = \frac{1}{m} \left( \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j} \right) \]

3. **Finding the Acceleration Vector**:
The acceleration \(\vec{a}\) is the time derivative of the velocity \(\vec{v}\):
\[ \vec{a}(t) = \frac{d\vec{v}(t)}{dt} \]
To differentiate \(\vec{v}(t)\):
\[ \vec{a}(t) = \frac{1}{m} \left( -\omega \sin(\omega t) \hat{i} + \omega \cos(\omega t) \hat{j} \right) \]

4. **Analyzing the Direction of Velocity and Acceleration**:
- The velocity vector \(\vec{v}(t)\) is in the direction of the momentum vector \(\vec{p}(t)\).
- The acceleration vector \(\vec{a}(t)\) is derived from the velocity vector and involves sine and cosine terms, indicating a circular motion.

5. **Checking Perpendicularity**:
To check if the velocity and acceleration vectors are perpendicular, we can take the dot product:
\[ \vec{v}(t) \cdot \vec{a}(t) = \left( \frac{1}{m} \left( \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j} \right) \right) \cdot \left( \frac{1}{m} \left( -\omega \sin(\omega t) \hat{i} + \omega \cos(\omega t) \hat{j} \right) \right) \]
This simplifies to:
\[ = \frac{\omega}{m^2} \left( \cos(\omega t)(\cos(\omega t)) + \sin(\omega t)(-\sin(\omega t)) \right) = \frac{\omega}{m^2} \left( \cos^2(\omega t) - \sin^2(\omega t) \right) \]
This expression is not necessarily zero, indicating that \(\vec{v}(t)\) and \(\vec{a}(t)\) are not perpendicular.

6. **Conclusion**:
- The velocity vector \(\vec{v}(t)\) is parallel to the momentum vector \(\vec{p}(t)\).
- The acceleration vector \(\vec{a}(t)\) is perpendicular to the velocity vector \(\vec{v}(t)\).

### Final Answer:
The correct statement is:
- **Velocity is parallel to momentum, and acceleration is perpendicular to velocity.**