Initially a body is at rest. If its acceleration is `5ms^(-2)` then the distance travelled in the `18^(th)` second is :-

Explanation

To solve the problem of finding the distance traveled by a body in the 18th second when it starts from rest with an acceleration of \(5 \, \text{m/s}^2\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial velocity (\(u\)) = 0 (since the body is at rest)
- Acceleration (\(a\)) = \(5 \, \text{m/s}^2\)
- Time interval for which we want to find the distance (\(n\)) = 18 seconds

2. **Use the Formula for Distance in the nth Second:**
The formula to calculate the distance traveled in the nth second is given by:
\[ s_n = u + \frac{1}{2} a (2n - 1) \]
where \(s_n\) is the distance traveled in the nth second.

3. **Substitute the Known Values:**
Since \(u = 0\) and \(a = 5 \, \text{m/s}^2\), we can substitute these values into the formula:
\[ s_{18} = 0 + \frac{1}{2} \times 5 \times (2 \times 18 - 1) \]

4. **Calculate the Expression Inside the Parenthesis:**
Calculate \(2 \times 18 - 1\):
\[ 2 \times 18 = 36 \]
\[ 36 - 1 = 35 \]

5. **Substitute Back into the Formula:**
Now substitute \(35\) back into the formula:
\[ s_{18} = \frac{1}{2} \times 5 \times 35 \]

6. **Perform the Multiplication:**
First calculate \(5 \times 35\):
\[ 5 \times 35 = 175 \]
Then calculate \(\frac{1}{2} \times 175\):
\[ \frac{1}{2} \times 175 = 87.5 \]

7. **Final Result:**
Therefore, the distance traveled in the 18th second is:
\[ s_{18} = 87.5 \, \text{meters} \]

### Final Answer:
The distance traveled in the 18th second is **87.5 meters**.

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