Velocity of a particle moving along a straight line at any time 't' is give by `V = cos((pi)/(3)t)` . Then the distance travelled by the particle in the first two seconds is :

Options

A. `(3)/(pi)[2-(sqrt(3))/(2)]`
B. `[2-(sqrt(3))/(2)]`
C. `(2)/(pi)[2-(sqrt(3))/(2)]`
D. `(4)/(pi)[2-(sqrt(3))/(2)]`

Explanation

To find the distance traveled by a particle moving along a straight line with a given velocity function \( V(t) = \cos\left(\frac{\pi}{3} t\right) \), we need to integrate the velocity function over the time interval from \( t = 0 \) to \( t = 2 \) seconds.

### Step-by-step Solution:

1. **Identify the velocity function**:
\[ V(t) = \cos\left(\frac{\pi}{3} t\right) \]

2. **Set up the integral for distance**:
The distance \( x \) traveled by the particle from time \( t = 0 \) to \( t = 2 \) seconds is given by:
\[ x = \int_{0}^{2} V(t) \, dt = \int_{0}^{2} \cos\left(\frac{\pi}{3} t\right) \, dt \]

3. **Integrate the function**:
To solve the integral, we can use the substitution method. Let:
\[ u = \frac{\pi}{3} t \quad \Rightarrow \quad du = \frac{\pi}{3} dt \quad \Rightarrow \quad dt = \frac{3}{\pi} du \]
Changing the limits:
- When \( t = 0 \), \( u = 0 \)
- When \( t = 2 \), \( u = \frac{2\pi}{3} \)

Now, we can rewrite the integral:
\[ x = \int_{0}^{\frac{2\pi}{3}} \cos(u) \cdot \frac{3}{\pi} \, du \]
\[ = \frac{3}{\pi} \int_{0}^{\frac{2\pi}{3}} \cos(u) \, du \]

4. **Evaluate the integral**:
The integral of \( \cos(u) \) is \( \sin(u) \):
\[ = \frac{3}{\pi} \left[ \sin(u) \right]_{0}^{\frac{2\pi}{3}} = \frac{3}{\pi} \left( \sin\left(\frac{2\pi}{3}\right) - \sin(0) \right) \]
\[ = \frac{3}{\pi} \left( \sin\left(\frac{2\pi}{3}\right) \right) \]
Since \( \sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \):
\[ = \frac{3}{\pi} \cdot \frac{\sqrt{3}}{2} \]

5. **Final distance calculation**:
\[ x = \frac{3\sqrt{3}}{2\pi} \]

### Conclusion:
The distance traveled by the particle in the first two seconds is:
\[ x = \frac{3\sqrt{3}}{2\pi} \]