The initial velocity of a body moving along a straight lines is `7m//s`. It has a uniform acceleration of `4 m//s^(2)` the distance covered by the body in the `5^(th)` second of its motion is-

Options

A. `25 m`
B. `35 m`
C. `50 m`
D. `85 m`

Explanation

To find the distance covered by the body in the 5th second of its motion, we can use the formula for the distance covered in the nth second of uniformly accelerated motion:

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

Where:
- \( S_n \) is the distance covered in the nth second,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( n \) is the nth second.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial velocity, \( u = 7 \, \text{m/s} \)
- Acceleration, \( a = 4 \, \text{m/s}^2 \)
- We need to find the distance covered in the 5th second, so \( n = 5 \).

2. **Substitute the values into the formula:**
\[ S_5 = u + \frac{a}{2} (2n - 1) \]
\[ S_5 = 7 + \frac{4}{2} (2 \times 5 - 1) \]

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

4. **Now substitute this back into the equation:**
\[ S_5 = 7 + \frac{4}{2} \times 9 \]
\[ S_5 = 7 + 2 \times 9 \]
\[ S_5 = 7 + 18 \]

5. **Finally, calculate the total distance covered in the 5th second:**
\[ S_5 = 25 \, \text{m} \]

### Final Answer:
The distance covered by the body in the 5th second of its motion is **25 meters**.

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