To calculate the displacement from the velocity-time graph between $t = 0$ and $t = 8$ seconds, we need to determine the area under the curve in this interval. Here's how we analyze this graph:

Step 1: Break the Graph into Shapes

The graph consists of two main sections between $t = 0$ and $t = 8$:

1. A triangle from $t = 0$ to $t = 5$.
2. A rectangle from $t = 5$ to $t = 8$.

Step 2: Calculate the Area of Each Section

The displacement is the total area under the curve:

1. Triangle from $t = 0$ to $t = 5$:

   • Base = $5 \, \text{s}$, Height = $5 \, \text{m/s}$
   • Area = $\frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 5 \times 5 = 12.5 \, \text{m}$

2. Rectangle from $t = 5$ to $t = 8$:

   • Base = $8 - 5 = 3 \, \text{s}$, Height = $5 \, \text{m/s}$
   • Area = $\text{Base} \times \text{Height} = 3 \times 5 = 15 \, \text{m}$

Step 3: Total Displacement

Total displacement = Area of triangle + Area of rectangle
$\text{Total displacement} = 12.5 \, \text{m} + 15 \, \text{m} = 27.5 \, \text{m}$

Answer

Since none of the provided options match $27.5$, it's possible there’s an issue with the question's data or options. However, this is the correct calculation for the given graph.

Would you like me to double-check or clarify any steps?

Here are 5 follow-up questions for additional learning:

1. How do you find displacement from a velocity-time graph if the curve dips below the time axis?
2. What is the physical significance of negative displacement?
3. How would the calculation change if the triangle’s peak was at $6 \, \text{s}$ instead of $5 \, \text{s}$?
4. What does the slope of a velocity-time graph represent?
5. How do you calculate total distance traveled when velocity changes direction?

Tip: Always verify if velocity-time graphs involve positive or negative regions, as this directly affects displacement and distance calculations.