The problem is asking to determine the total distance traveled by the object based on the velocity-time (v-t) graph shown. Let me analyze the graph first.

Graph Interpretation:

• The y-axis represents velocity $v$ in $m/s$, and the x-axis represents time $t$ in seconds.
• The graph has three distinct parts:
  1. From $t = 0$ to $t = 4$ seconds: The velocity increases linearly from 0 to 4 m/s.
  2. From $t = 4$ to $t = 8$ seconds: The velocity stays constant at 4 m/s.
  3. From $t = 8$ to $t = 10$ seconds: The velocity decreases linearly from 4 m/s to -4 m/s.

Distance Traveled Calculation:

To calculate the total distance traveled, we need to compute the area under the velocity-time graph. The total distance is the sum of the absolute values of the areas (because the velocity goes negative in the last section, indicating a change in direction).

1. From $t = 0$ to $t = 4$ seconds:

   • This forms a triangle with a base of $4 \, \text{s}$ and a height of $4 \, \text{m/s}$.
   • Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \, \text{m}$.

2. From $t = 4$ to $t = 8$ seconds:

   • This forms a rectangle with a base of $4 \, \text{s}$ and a height of $4 \, \text{m/s}$.
   • Area of rectangle $= \text{base} \times \text{height} = 4 \times 4 = 16 \, \text{m}$.

3. From $t = 8$ to $t = 10$ seconds:

   • This forms a triangle with a base of $2 \, \text{s}$ and a height of $4 \, \text{m/s}$ (but negative velocity).
   • Area of triangle $= \frac{1}{2} \times 2 \times 4 = 4 \, \text{m}$.

Thus, the total distance traveled is the sum of the absolute areas: $\text{Total distance} = 8 + 16 + 4 = 28 \, \text{m}.$

Let me know if you want further explanation or details on the process!

Relative Questions:

1. How does the negative velocity affect the total displacement?
2. What is the difference between distance and displacement in motion analysis?
3. Can you explain how acceleration would be calculated from this graph?
4. How would the graph look if the object returned to the starting position?
5. What is the significance of calculating the area under a v-t graph?

Tip:

Always remember that in a velocity-time graph, the area under the curve represents the object's displacement, while the total distance considers the absolute value of that area.