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. Here's a step-by-step breakdown:

### Step 1: Understand the momentum vector
The momentum vector is given as:
\[ \vec{p}(t) = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j} \]
where \(\omega\) is a constant and \(t\) is time.

### Step 2: Relate momentum to velocity
The momentum \(\vec{p}\) is related to velocity \(\vec{v}\) by the equation:
\[ \vec{p} = m \vec{v} \]
From this, we can express the velocity vector as:
\[ \vec{v}(t) = \frac{\vec{p}(t)}{m} = \frac{1}{m} \left( \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j} \right) \]

### Step 3: Differentiate the velocity to find acceleration
To find the acceleration \(\vec{a}(t)\), we differentiate the velocity vector with respect to time:
\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{1}{m} \frac{d}{dt} \left( \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j} \right) \]
Using the chain rule:
\[ \vec{a}(t) = \frac{1}{m} \left( -\omega \sin(\omega t) \hat{i} + \omega \cos(\omega t) \hat{j} \right) \]
This simplifies to:
\[ \vec{a}(t) = \frac{\omega}{m} \left( -\sin(\omega t) \hat{i} + \cos(\omega t) \hat{j} \right) \]

### Step 4: Check the relationship between \(\vec{v}\) and \(\vec{p}\)
From the earlier steps, we have:
- \(\vec{v}(t) = \frac{1}{m} \left( \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j} \right)\)
- \(\vec{p}(t) = m \vec{v}(t)\)

This indicates that \(\vec{v}(t)\) is parallel to \(\vec{p}(t)\).

### Step 5: Check the relationship between \(\vec{a}\) and \(\vec{p}\)
Now, we need to check if \(\vec{a}(t)\) is perpendicular to \(\vec{p}(t)\):
\[ \vec{a}(t) \cdot \vec{p}(t) = \left( -\frac{\omega}{m} \sin(\omega t) \hat{i} + \frac{\omega}{m} \cos(\omega t) \hat{j} \right) \cdot \left( \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j} \right) \]
Calculating the dot product:
\[ = -\frac{\omega}{m} \sin(\omega t) \cos(\omega t) + \frac{\omega}{m} \cos(\omega t) \sin(\omega t) = 0 \]
Since the dot product is zero, \(\vec{a}(t)\) is perpendicular to \(\vec{p}(t)\).

### Conclusion
From the analysis:
- \(\vec{v}(t)\) is parallel to \(\vec{p}(t)\).
- \(\vec{a}(t)\) is perpendicular to \(\vec{p}(t)\).

Thus, the correct statement is:
- **Option 4**: \(\vec{v}\) is parallel to \(\vec{p}\) and \(\vec{a}\) is perpendicular to \(\vec{p}\).