To find the distance traveled over 32 seconds from the given speed-time graph, we need to calculate the area under the graph.

Step-by-Step Solution:

The graph shows three sections:

1. First Section (0 to 12 seconds): A triangular area where the speed increases linearly from 0 m/s to 20 m/s.
2. Second Section (12 to 32 seconds): A rectangular area where the speed remains constant at 20 m/s.

1. Area of the First Section (Triangle)

• Base (time interval) = 12 seconds
• Height (speed) = 20 m/s
• Area of a triangle = $\frac{1}{2} \times \text{base} \times \text{height}$

$\text{Area}_{\text{triangle}} = \frac{1}{2} \times 12 \times 20 = 120 \text{ m}$

2. Area of the Second Section (Rectangle)

• Length (time interval) = 32 - 12 = 20 seconds
• Width (speed) = 20 m/s
• Area of a rectangle = length $\times$ width

$\text{Area}_{\text{rectangle}} = 20 \times 20 = 400 \text{ m}$

Total Distance

Now, add the areas of both sections:

$\text{Total Distance} = \text{Area}_{\text{triangle}} + \text{Area}_{\text{rectangle}} = 120 + 400 = 520 \text{ m}$

Answer:

The total distance traveled over 32 seconds is 520 meters.

Would you like more details, or do you have any questions?

Here are some related questions that might enhance your understanding:

1. What is the formula for calculating distance from a speed-time graph?
2. How would the result change if the constant speed section lasted longer than 20 seconds?
3. How would you find the average speed of the particle over the entire 32-second period?
4. What would be the distance if the initial slope had been steeper?
5. How can we determine acceleration from the graph?

Tip: Remember, the area under a speed-time graph always represents the distance traveled.