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From the diagram and notes provided, it appears that you are working on a velocity-time graph involving acceleration and deceleration phases. The distance traveled in such cases can be determined by calculating the area under the velocity-time graph.

Looking at your sketch:

1. **Constant speed phase (rectangle area)**: The velocity is constant (10 m/s) for 30 seconds.
   \[
   \text{Area} = \text{base} \times \text{height} = 10 \times 30 = 300 \, \text{m}
   \]

2. **Acceleration phase (triangle area)**: The velocity increases linearly from 10 m/s to 40 m/s in 10 seconds.
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times (40 - 10) = 150 \, \text{m}
   \]

3. **Deceleration phase (triangle area)**: The velocity decreases from 50 m/s to 0 m/s over 60 seconds.
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 50 \times 60 = 1500 \, \text{m}
   \]

Summing all the areas:
\[
\text{Total Distance} = 300 + 150 + 1500 = 1950 \, \text{m}
\]

The total distance traveled is approximately **1950 meters**.

Do you want more details or clarifications on any step?

Here are 5 related questions for further exploration:
1. How do we interpret the area under a velocity-time graph?
2. How do you calculate the area of a trapezoid in velocity-time graphs?
3. What happens to the distance if the acceleration is doubled?
4. How do we calculate average velocity from a velocity-time graph?
5. What is the importance of knowing the acceleration and deceleration phases in real-world scenarios?

**Tip**: When working with velocity-time graphs, always remember that the area under the curve represents the total distance traveled!

Learn how to calculate the total distance traveled using a velocity-time graph with acceleration and deceleration phases. The problem involves calculating the areas under the graph to find the distance. Suitable for Grades 10-12 students.

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