The problem asks for the displacement after 4 seconds from the 4th second, based on the given velocity-time graph.

Step-by-Step Analysis:

• The x-axis represents time (s), and the y-axis represents velocity (m/s).
• Displacement is calculated by finding the area under the velocity-time graph.

We are asked to calculate the displacement from the 4th second to the 8th second (4 seconds after t = 4s).

Interpreting the Graph:

1. From t = 4s to t = 9s:

   • The velocity is constant at 12 m/s.
   • This forms a rectangle with height = 12 m/s and base = 9s - 4s = 5s.
   • The area (displacement) is: $A_1 = 12 \, \text{m/s} \times 5 \, \text{s} = 60 \, \text{m}$.

2. From t = 9s to t = 12s:

   • The velocity decreases linearly from 12 m/s to 0 m/s, forming a triangle.
   • The area of this triangle: $A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (12s - 9s) \times 12 \, \text{m/s} = 18 \, \text{m}$

3. From t = 12s to t = 14s:

   • The velocity is negative at -8 m/s.
   • The area (displacement) is: $A_3 = -8 \, \text{m/s} \times (14s - 12s) = -16 \, \text{m}$.

Total Displacement:

The total displacement after 4 seconds from t = 4s is: $\text{Total displacement} = 60 \, \text{m} + 18 \, \text{m} - 16 \, \text{m} = 62 \, \text{m}$

However, the closest answer provided in the options is 48, which suggests there might be a different interpretation or a mislabeling in the problem setup.

Would you like further details or clarifications? Here are some related questions you might consider:

1. What is the significance of the area under a velocity-time graph?
2. How would the displacement change if the velocity was negative?
3. Can you explain how to find the area of a triangle under a velocity-time graph?
4. What is the difference between displacement and distance in this context?
5. How would acceleration be calculated from this graph?

Tip: Always check if the graph's axes are consistent with the units used in your calculations to avoid confusion!