A body starting from rest and has uniform acceleration
`8 m//s^(2)` . The distance travelled by it in` 5^(th)` second will be

Options

A. 36 m
B. 40 m
C. 100 m
D. 200 m

Answer

To solve the problem of finding the distance traveled by a body in the 5th second when it starts from rest with a uniform acceleration of \(8 \, \text{m/s}^2\), we can use the formula for the distance traveled in the nth second of motion:

\[ d_n = u + \frac{1}{2} a (2n - 1) \]

where:
- \(d_n\) is the distance traveled in the nth second,
- \(u\) is the initial velocity,
- \(a\) is the acceleration,
- \(n\) is the specific second for which we want to find the distance.

### Step-by-Step Solution:

1. **Identify the given values**:
- Initial velocity \(u = 0 \, \text{m/s}\) (since the body starts from rest)
- Acceleration \(a = 8 \, \text{m/s}^2\)
- We want to find the distance traveled in the 5th second, so \(n = 5\).

2. **Substitute the values into the formula**:
\[ d_5 = 0 + \frac{1}{2} \times 8 \times (2 \times 5 - 1) \]

3. **Calculate the expression inside the parentheses**:
\[ 2 \times 5 - 1 = 10 - 1 = 9 \]

4. **Now substitute this back into the equation**:
\[ d_5 = \frac{1}{2} \times 8 \times 9 \]

5. **Calculate \(\frac{1}{2} \times 8\)**:
\[ \frac{1}{2} \times 8 = 4 \]

6. **Now multiply by 9**:
\[ d_5 = 4 \times 9 = 36 \, \text{meters} \]

### Final Answer:
The distance traveled by the body in the 5th second is \(36 \, \text{meters}\).

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