A body starts with an initial velocity of 10 ms and acceleration 5 ms. Find the distance covered by it in 5 s.

Explanation

To solve the problem of finding the distance covered by a body with an initial velocity and constant acceleration, we can use the following kinematic equation:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( s \) = distance covered
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

### Step-by-step Solution:

1. **Identify the given values:**
- Initial velocity (\( u \)) = 10 m/s
- Acceleration (\( a \)) = 5 m/s²
- Time (\( t \)) = 5 s

2. **Substitute the values into the equation:**
\[ s = ut + \frac{1}{2} a t^2 \]
Substitute \( u = 10 \, \text{m/s} \), \( a = 5 \, \text{m/s}^2 \), and \( t = 5 \, \text{s} \):
\[ s = (10 \, \text{m/s} \times 5 \, \text{s}) + \frac{1}{2} \times (5 \, \text{m/s}^2) \times (5 \, \text{s})^2 \]

3. **Calculate the first term:**
\[ 10 \, \text{m/s} \times 5 \, \text{s} = 50 \, \text{m} \]

4. **Calculate the second term:**
\[ \frac{1}{2} \times 5 \, \text{m/s}^2 \times 25 \, \text{s}^2 = \frac{1}{2} \times 5 \times 25 = \frac{125}{2} = 62.5 \, \text{m} \]

5. **Add both terms to find the total distance:**
\[ s = 50 \, \text{m} + 62.5 \, \text{m} = 112.5 \, \text{m} \]

### Final Answer:
The distance covered by the body in 5 seconds is **112.5 meters**.

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