To solve this problem, we need to determine the distances between the students based on their relative positions. Let’s represent their positions in a coordinate system.
-
Position of Alexandra: Let’s place Alexandra at the origin, \(A = (0, 0)\).
-
Position of Roy: According to the problem, Roy is 4.4 meters directly behind Alexandra. Assuming that "behind" refers to the negative direction on the y-axis, we can place Roy at: \[ R = (0, -4.4) \]
-
Position of Laura: Roy is also 2.9 meters directly left of Laura. If "left" means to the negative direction on the x-axis, then we can determine Laura's coordinates. Since Roy is 2.9 meters left of Laura, we can express Laura's position as: \[ L = (2.9, -4.4) \]
Now we can calculate the distances the paper airplane travels:
Distance from Roy to Alexandra:
The distance \(d_{RA}\) from Roy to Alexandra can be calculated using the distance formula: \[ d_{RA} = \sqrt{(0 - 0)^2 + (0 + 4.4)^2} = \sqrt{(0)^2 + (4.4)^2} = \sqrt{19.36} \approx 4.4 \text{ meters} \]
Distance from Alexandra to Laura:
Next, we calculate the distance \(d_{AL}\) from Alexandra to Laura: \[ d_{AL} = \sqrt{(2.9 - 0)^2 + (-4.4 - 0)^2} = \sqrt{(2.9)^2 + (-4.4)^2} = \sqrt{8.41 + 19.36} = \sqrt{27.77} \approx 5.3 \text{ meters} \]
Distance from Laura back to Roy:
Lastly, we find the distance \(d_{LR}\) from Laura back to Roy: \[ d_{LR} = \sqrt{(0 - 2.9)^2 + (-4.4 + 4.4)^2} = \sqrt{(-2.9)^2 + (0)^2} = \sqrt{8.41} \approx 2.9 \text{ meters} \]
Total Distance:
Now we can sum these distances together to find the total distance that the paper airplane has traveled: \[ \text{Total distance} = d_{RA} + d_{AL} + d_{LR} \approx 4.4 + 5.3 + 2.9 = 12.6 \text{ meters} \]
Therefore, the total distance the paper airplane traveled is approximately 12.6 meters.