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 displacement during a specific second of motion. The distance covered in the nth second can be calculated using the formula:

\[ S_n = U + \frac{1}{2} A (2n - 1) \]

Where:
- \( S_n \) is the distance covered in the nth second,
- \( U \) is the initial velocity,
- \( A \) is the acceleration,
- \( n \) is the specific second (in this case, \( n = 5 \)).

### Step 1: Identify the given values
- Initial velocity, \( U = 7 \, \text{m/s} \)
- Acceleration, \( A = 4 \, \text{m/s}^2 \)
- Time for the 5th second, \( n = 5 \)

### Step 2: Substitute the values into the formula
Now, substitute the values into the formula for \( S_n \):

\[ S_5 = U + \frac{1}{2} A (2 \cdot 5 - 1) \]

### Step 3: Calculate \( S_5 \)
First, calculate \( 2n - 1 \):

\[ 2 \cdot 5 - 1 = 10 - 1 = 9 \]

Now substitute this back into the equation:

\[ S_5 = 7 + \frac{1}{2} \cdot 4 \cdot 9 \]

### Step 4: Simplify the equation
Calculate \( \frac{1}{2} \cdot 4 \cdot 9 \):

\[ \frac{1}{2} \cdot 4 = 2 \]
\[ 2 \cdot 9 = 18 \]

Now substitute this value back into the equation:

\[ S_5 = 7 + 18 = 25 \, \text{meters} \]

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

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