To determine the total distance traveled by the object from 0 to 40 seconds using the velocity vs. time graph, we need to calculate the area under the graph within that time interval.

The graph appears to be made up of three distinct linear sections:

1. An upward-sloping line from 0 to approximately 12 seconds, reaching a peak velocity.
2. A horizontal line representing constant velocity from about 12 seconds to around 22 seconds.
3. A downward-sloping line back to zero from about 22 seconds to 40 seconds.

To find the total distance:

1. Calculate the area of each section:

   • First Section (0 to 12 seconds): This is a right triangle.
   • Second Section (12 to 22 seconds): This is a rectangle.
   • Third Section (22 to 40 seconds): This is another right triangle.

2. Compute each area:

   • First Triangle Area (0 to 12 s):
     Base = 12 s, Height = 25 m/s
     Area = $\frac{1}{2} \times 12 \times 25 = 150 \, \text{m}$

   • Rectangle Area (12 to 22 s):
     Width = 22 s - 12 s = 10 s, Height = 25 m/s
     Area = $10 \times 25 = 250 \, \text{m}$

   • Second Triangle Area (22 to 40 s):
     Base = 40 s - 22 s = 18 s, Height = 25 m/s
     Area = $\frac{1}{2} \times 18 \times 25 = 225 \, \text{m}$

3. Sum the areas to find the total distance: $\text{Total Distance} = 150 + 250 + 225 = 625 \, \text{m}$

So, the total distance traveled by the object from 0 to 40 seconds is 625 meters.

Would you like more details or have any questions?

Here are some related questions that could further expand your understanding:

1. How do you calculate the displacement if the object changes direction during the motion?
2. What does the slope of the velocity-time graph represent?
3. How would the graph look if the object were accelerating uniformly?
4. What is the difference between speed and velocity in terms of motion?
5. How would you determine the object's acceleration at any point using this graph?

Tip: When interpreting graphs, always consider the shape and area under the curve to determine quantities like distance and displacement.