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The task asks to determine the distance traveled by a body in the time interval from \( t = 0 \) to \( t = 5 \) seconds, based on the graph of its velocity \( v_x \) over time \( t \).

### Key Concepts:
- The graph provided is a velocity-time graph, and the area under the curve within a certain time interval represents the displacement (if areas are taken with signs) or the total distance traveled (if areas are taken as absolute values).
- The distance is obtained by summing the absolute values of the areas under the velocity curve over the time interval.

### Step-by-Step Solution:
From \( t = 0 \) to \( t = 5 \), the velocity graph can be broken into distinct shapes (rectangles and triangles), and we will calculate the area of each.

1. **From \( t = 0 \) to \( t = 2 \)**:
   - This forms a triangle with a base of \( 2 \) seconds and a height of \( 10 \, \text{m/s} \).
   - Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 10 = 10 \, \text{m} \).

2. **From \( t = 2 \) to \( t = 4 \)**:
   - This forms a triangle with a base of \( 2 \) seconds and a height of \( -10 \, \text{m/s} \) (negative because it's below the time axis).
   - The area (displacement) = \( \frac{1}{2} \times 2 \times 10 = 10 \, \text{m} \) (absolute value is taken to calculate distance).

3. **From \( t = 4 \) to \( t = 5 \)**:
   - This forms a small rectangle with a base of \( 1 \) second and a height of \( 5 \, \text{m/s} \).
   - Area of rectangle = \( 1 \times 5 = 5 \, \text{m} \).

### Total Distance:
The total distance traveled is the sum of the absolute values of the areas:
\[
\text{Distance} = 10 \, \text{m} + 10 \, \text{m} + 5 \, \text{m} = 25 \, \text{m}
\]

Thus, the distance traveled from \( t = 0 \) to \( t = 5 \) seconds is **25 meters**.

Would you like any further explanation or details on any of the steps?

Here are 5 related questions:
1. How would the result change if the velocity graph included negative areas (representing opposite direction)?
2. What would be the displacement from \( t = 0 \) to \( t = 5 \) seconds?
3. How is displacement different from total distance traveled?
4. How do you calculate the area under a curve in more complex velocity-time graphs?
5. How does constant acceleration affect the velocity-time graph?

**Tip:** Always take care to differentiate between distance and displacement when dealing with velocity-time graphs. Distance is always positive, but displacement can be negative.

На рисунке приведен график зависимости проекции vx скорости тела от времени t. Определите путь, пройденный телом в интервале времени от 0 до 5 с. Ответ приведите в метрах.

Determine the distance traveled by a body in the time interval from 0 to 5 seconds using a velocity-time graph. This problem involves calculating the area under the curve, representing the total distance covered. Suitable for students in Grades 10-12.

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