To solve the problem of the body moving with varying velocities, we can break it down into three segments based on the time intervals provided. We will calculate the distance covered in each segment and then sum them up to find the total distance covered after 12 seconds.
### Step-by-Step Solution
1. **Segment 1 (0 to 5 seconds)**:
- The body moves with a constant velocity of \(2 \, \text{m/s}\) for \(5 \, \text{s}\).
- Distance covered in this segment:
\[
\text{Distance}_1 = \text{velocity} \times \text{time} = 2 \, \text{m/s} \times 5 \, \text{s} = 10 \, \text{m}
\]
2. **Segment 2 (5 to 10 seconds)**:
- The velocity increases uniformly from \(2 \, \text{m/s}\) to \(10 \, \text{m/s}\) over \(5 \, \text{s}\).
- The average velocity during this segment can be calculated as:
\[
\text{Average Velocity} = \frac{\text{Initial Velocity} + \text{Final Velocity}}{2} = \frac{2 \, \text{m/s} + 10 \, \text{m/s}}{2} = 6 \, \text{m/s}
\]
- Distance covered in this segment:
\[
\text{Distance}_2 = \text{Average Velocity} \times \text{time} = 6 \, \text{m/s} \times 5 \, \text{s} = 30 \, \text{m}
\]
3. **Segment 3 (10 to 15 seconds)**:
- The body decelerates uniformly from \(10 \, \text{m/s}\) to \(0 \, \text{m/s}\) over the next \(5 \, \text{s}\).
- The average velocity during this segment is:
\[
\text{Average Velocity} = \frac{10 \, \text{m/s} + 0 \, \text{m/s}}{2} = 5 \, \text{m/s}
\]
- Distance covered in this segment:
\[
\text{Distance}_3 = \text{Average Velocity} \times \text{time} = 5 \, \text{m/s} \times 5 \, \text{s} = 25 \, \text{m}
\]
4. **Total Distance Covered**:
- Now, we sum up the distances covered in all three segments:
\[
\text{Total Distance} = \text{Distance}_1 + \text{Distance}_2 + \text{Distance}_3
\]
\[
\text{Total Distance} = 10 \, \text{m} + 30 \, \text{m} + 25 \, \text{m} = 65 \, \text{m}
\]
### Final Answer
The total distance covered by the body after 12 seconds is **65 meters**.
