A particle is moving along a straight line along x-zxis with an initial velocity of `2 m//s` towards positive x-axis. A constant acceleration of `0.5 m//s^(2)` towards negative x-axis starts acting on particle at `t=0`. Find velocity (in `m//s`) of particle at `t=2 s`.

Explanation

To solve the problem, we will use the equation of motion that relates initial velocity, acceleration, time, and final velocity. The equation is:

\[ v = u + at \]

Where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity, \( u = 2 \, \text{m/s} \) (towards positive x-axis)
- Acceleration, \( a = -0.5 \, \text{m/s}^2 \) (towards negative x-axis)
- Time, \( t = 2 \, \text{s} \)

2. **Substitute the Values into the Equation:**
We will substitute the values of \( u \), \( a \), and \( t \) into the equation \( v = u + at \):

\[ v = 2 \, \text{m/s} + (-0.5 \, \text{m/s}^2) \times (2 \, \text{s}) \]

3. **Calculate the Acceleration Component:**
Calculate \( at \):

\[ at = -0.5 \, \text{m/s}^2 \times 2 \, \text{s} = -1 \, \text{m/s} \]

4. **Calculate the Final Velocity:**
Now substitute \( at \) back into the equation for \( v \):

\[ v = 2 \, \text{m/s} - 1 \, \text{m/s} = 1 \, \text{m/s} \]

5. **Conclusion:**
The final velocity of the particle at \( t = 2 \, \text{s} \) is:

\[ v = 1 \, \text{m/s} \]