To find how far apart the two airplanes are after 3.3 hours, we will analyze their respective positions using vector components based on their velocities and headings.
Step 1: Determine the velocities in component form
- For the first airplane:
- Velocity \( V_1 = 730 \) m/h
- Heading = 53.6 degrees
The components of the velocity can be calculated using trigonometric functions: \[ V_{1x} = V_1 \cdot \cos(53.6^\circ) \] \[ V_{1y} = V_1 \cdot \sin(53.6^\circ) \]
- For the second airplane:
- Velocity \( V_2 = 570 \) m/h
- Heading = 88 degrees
Similarly, we calculate the components: \[ V_{2x} = V_2 \cdot \cos(88^\circ) \] \[ V_{2y} = V_2 \cdot \sin(88^\circ) \]
Step 2: Calculate the components
Using a calculator:
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For the first airplane: \[ V_{1x} = 730 \cdot \cos(53.6^\circ) \approx 730 \cdot 0.599 = 436.27 \text{ m/h} \] \[ V_{1y} = 730 \cdot \sin(53.6^\circ) \approx 730 \cdot 0.800 = 584.00 \text{ m/h} \]
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For the second airplane: \[ V_{2x} = 570 \cdot \cos(88^\circ) \approx 570 \cdot 0.0349 = 19.89 \text{ m/h} \] \[ V_{2y} = 570 \cdot \sin(88^\circ) \approx 570 \cdot 0.999 = 569.43 \text{ m/h} \]
Step 3: Calculate the positions after 3.3 hours
Next, determine the total displacement for each airplane after \( t = 3.3 \) hours.
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For the first airplane:
\[ x_1 = V_{1x} \cdot t = 436.27 \cdot 3.3 \approx 1440.78 \text{ m} \] \[ y_1 = V_{1y} \cdot t = 584.00 \cdot 3.3 \approx 1927.20 \text{ m} \] -
For the second airplane:
\[ x_2 = V_{2x} \cdot t = 19.89 \cdot 3.3 \approx 65.96 \text{ m} \] \[ y_2 = V_{2y} \cdot t = 569.43 \cdot 3.3 \approx 1888.80 \text{ m} \]
Step 4: Calculate the distance between the two airplanes
Now, we find the distance \( d \) between the two final positions \( (x_1, y_1) \) and \( (x_2, y_2) \) using the distance formula: \[ d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]
Substituting the calculated values: \[ d = \sqrt{(1440.78 - 65.96)^2 + (1927.20 - 1888.80)^2} \] \[ d = \sqrt{(1374.82)^2 + (38.40)^2} \] Calculating the squares: \[ d = \sqrt{1897910.1284 + 1476.096} \] \[ d = \sqrt{1899386.2244} \approx 1379.54 \text{ m} \]
Thus, the distance between the two airplanes after 3.3 hours is approximately 1379.54 m.
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