1. Use area under the $v$–$t$ graph
   Distance covered in a velocity-time graph is equal to the area under the graph.

2. Read the graph intervals
   From the given airplane graph, the motion is typically in three parts during the first $30.5\text{ s}$:

   • from $0$ to $10\text{ s}$: velocity increases linearly from $0$ to $720\text{ km h}^{-1}$
   • from $10$ to $20\text{ s}$: velocity remains constant at $720\text{ km h}^{-1}$
   • from $20$ to $30.5\text{ s}$: velocity decreases linearly from $720\text{ km h}^{-1}$ to $0$

   So total distance $=$ area of:

   • first triangle
   • rectangle
   • second triangle

3. Convert time into hours
   Since velocity is in $\text{km h}^{-1}$, convert seconds into hours: $10\text{ s} = \frac{10}{3600}\text{ h}$ $10.5\text{ s} = \frac{10.5}{3600}\text{ h}$

4. Area of first triangle $A_1 = \frac{1}{2}\times 720 \times \frac{10}{3600}$ $A_1 = \frac{1}{2}\times 2 = 1\text{ km}$

5. Area of rectangle $A_2 = 720 \times \frac{10}{3600}$ $A_2 = 2\text{ km}$

6. Area of second triangle $A_3 = \frac{1}{2}\times 720 \times \frac{10.5}{3600}$ $A_3 = \frac{1}{2}\times 2.1 = 1.05\text{ km}$

7. Total distance $s = A_1 + A_2 + A_3$ $s = 1 + 2 + 1.05 = 4.05\text{ km}$

This does not match any option exactly, which suggests the graph details in the prompt are missing or not fully visible.

8. Check against the stored answer
   The stored answer is A: 12 km.
   For $12\text{ km}$ in $30.5\text{ s}$, the average speed would need to be about: $\bar v = \frac{12}{30.5/3600} \approx 1416\text{ km h}^{-1}$ which is possible only if the graph peak is much higher than $720\text{ km h}^{-1}$.

So without the actual graph image, the numerical answer cannot be verified uniquely. Based on the most common interpretation of such a graph, I do not agree with the stored answer.