A particle is moving with a position vector, `vec(r)=[a_(0) sin (2pi t) hat(i)+a_(0) cos (2pi t) hat(j)]`. Then -

Options

A. Magnitude of displacement of the particle between time `t=4` sec and `t=6` sec is zero
B. Distance travelled by the particle in 1 sec is `2pia_(0)`
C. The speed of particle in the whole motion is constant and equal to `2pia_(0)`.
D. none of these

Answer

To solve the given problem, we will analyze the motion of the particle based on its position vector and determine the validity of the provided options step by step.

### Given:
The position vector of the particle is:
\[ \vec{r}(t) = a_0 \sin(2\pi t) \hat{i} + a_0 \cos(2\pi t) \hat{j} \]

### Step 1: Identify the components of the position vector
From the position vector, we can identify the components:
- \( x(t) = a_0 \sin(2\pi t) \)
- \( y(t) = a_0 \cos(2\pi t) \)

### Step 2: Determine the path of the particle
To understand the motion, we can square and add the components:
\[ x^2 + y^2 = (a_0 \sin(2\pi t))^2 + (a_0 \cos(2\pi t))^2 \]
\[ = a_0^2 (\sin^2(2\pi t) + \cos^2(2\pi t)) = a_0^2 \]
This indicates that the particle moves in a circular path with radius \( a_0 \).

### Step 3: Analyze the displacement between \( t = 4 \) s and \( t = 6 \) s
Calculate the position vector at \( t = 4 \) s:
\[ \vec{r}(4) = a_0 \sin(8\pi) \hat{i} + a_0 \cos(8\pi) \hat{j} = a_0 (0) \hat{i} + a_0 (1) \hat{j} = a_0 \hat{j} \]

Calculate the position vector at \( t = 6 \) s:
\[ \vec{r}(6) = a_0 \sin(12\pi) \hat{i} + a_0 \cos(12\pi) \hat{j} = a_0 (0) \hat{i} + a_0 (1) \hat{j} = a_0 \hat{j} \]

The displacement between \( t = 4 \) s and \( t = 6 \) s is:
\[ \Delta \vec{r} = \vec{r}(6) - \vec{r}(4) = a_0 \hat{j} - a_0 \hat{j} = 0 \]
Thus, the magnitude of displacement is \( 0 \).

### Step 4: Calculate the distance traveled in one second
The particle completes one full cycle from \( t = 0 \) s to \( t = 1 \) s. The distance traveled in one complete cycle (circumference of the circle) is:
\[ \text{Distance} = 2\pi r = 2\pi a_0 \]

### Step 5: Determine the speed of the particle
The velocity vector can be derived from the position vector:
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(a_0 \sin(2\pi t) \hat{i} + a_0 \cos(2\pi t) \hat{j}) \]
\[ = a_0 (2\pi \cos(2\pi t) \hat{i} - 2\pi \sin(2\pi t) \hat{j}) \]
The magnitude of the velocity is:
\[ |\vec{v}| = \sqrt{(2\pi a_0 \cos(2\pi t))^2 + (-2\pi a_0 \sin(2\pi t))^2} \]
\[ = 2\pi a_0 \sqrt{\cos^2(2\pi t) + \sin^2(2\pi t)} = 2\pi a_0 \]
Thus, the speed of the particle is constant and equal to \( 2\pi a_0 \).

### Conclusion:
1. The magnitude of displacement between \( t = 4 \) s and \( t = 6 \) s is \( 0 \) (Correct).
2. The distance traveled by the particle in one second is \( 2\pi a_0 \) (Correct).
3. The speed of the particle in the whole motion is constant and equal to \( 2\pi a_0 \) (Correct).

### Final Answer:
All options are correct.