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!