A particle starts moving with acceleration `2 m//s^(2)`. Distance travelled by it in `5^(th)` half second is

Options

A. 1.25 m
B. 2.25 m
C. 6.25 m
D. 30.25 m

Answer

To solve this problem, we need to find the distance traveled by the particle in the 5th half-second interval. Let's break down the problem step by step.

### Step-by-Step Solution:

1. **Identify the given data:**
- Initial velocity (\(u\)) = 0 (since the particle starts from rest)
- Acceleration (\(a\)) = \(2 \, \text{m/s}^2\)
- We need to find the distance traveled in the 5th half-second.

2. **Calculate the total distance traveled by the particle up to \(2.5\) seconds:**
- Use the second equation of motion:
\[ S = ut + \frac{1}{2}at^2 \]
- Since \(u = 0\), the equation simplifies to:
\[ S = \frac{1}{2}at^2 \]
- Substitute \(a = 2 \, \text{m/s}^2\) and \(t = 2.5 \, \text{s}\):
\[ S_{2.5} = \frac{1}{2} \times 2 \times (2.5)^2 = (2.5)^2 = 6.25 \, \text{m} \]

3. **Calculate the total distance traveled by the particle up to \(2\) seconds:**
- Use the same equation:
\[ S_{2} = \frac{1}{2} \times 2 \times (2)^2 = 2 \times 2 = 4 \, \text{m} \]

4. **Find the distance traveled in the 5th half-second interval:**
- The 5th half-second interval is from \(2\) seconds to \(2.5\) seconds.
- The distance traveled in this interval is the difference between the distances calculated for \(2.5\) seconds and \(2\) seconds:
\[ S_{\text{5th half-second}} = S_{2.5} - S_{2} = 6.25 \, \text{m} - 4 \, \text{m} = 2.25 \, \text{m} \]

### Final Answer:
The distance traveled by the particle in the 5th half-second is \(2.25 \, \text{m}\).