13. Two airplanes depart from the same place at 2:00 p.m. One plane flies south at speed of 376 km/hr, and the other flies west at a speed of 648 km/hr. How far apart are the airplanes at 5:30 p.m.? Draw a picture and explain. How
13. Two airplanes depart from the same place at 2:00 p.m. One plane flies south at speed of 376 km/hr, and the other flies west at a speed of 648 km/hr. How far apart are the airplanes at 5:30 p.m.? Draw a picture and explain. How
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![**Problem 13: Airplane Distance Calculation**
Two airplanes depart from the same place at 2:00 p.m. One plane flies south at a speed of 376 km/hr, and the other flies west at a speed of 648 km/hr. How far apart are the airplanes at 5:30 p.m.? Draw a picture and explain.
**Explanation:**
To solve this problem, we need to find the distance between two airplanes moving in perpendicular directions after a certain period. This can be visualized as a right triangle, where each leg of the triangle represents the distance traveled by each airplane, and the hypotenuse represents the distance between the two airplanes.
- **Time Calculation:**
From 2:00 p.m. to 5:30 p.m. is 3.5 hours.
- **Distance Calculation:**
- The airplane flying south travels at 376 km/hr:
\[
\text{Distance South} = 376 \, \text{km/hr} \times 3.5 \, \text{hours} = 1316 \, \text{km}
\]
- The airplane flying west travels at 648 km/hr:
\[
\text{Distance West} = 648 \, \text{km/hr} \times 3.5 \, \text{hours} = 2268 \, \text{km}
\]
- **Using the Pythagorean Theorem:**
With these distances, use the Pythagorean theorem to find the hypotenuse (distance apart):
\[
\text{Distance Apart} = \sqrt{(1316^2 + 2268^2)}
\]
\[
\text{Distance Apart} = \sqrt{(1731856 + 5143824)} = \sqrt{6875680} \approx 2622 \, \text{km}
\]
Thus, the airplanes are approximately 2622 kilometers apart at 5:30 p.m.
**Diagram Explanation:**
- Imagine a right triangle:
- The vertical leg represents the path of the airplane flying south (1316 km).
- The horizontal leg represents the path of the airplane flying west (2268 km).
- The hypotenuse is the direct distance between the airplanes (approximately 2622 km).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79d4d229-649f-4192-a00b-385a235528b8%2F91e3eab6-fffb-434b-94aa-388f8f647b2e%2Fqmgjimi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 13: Airplane Distance Calculation**
Two airplanes depart from the same place at 2:00 p.m. One plane flies south at a speed of 376 km/hr, and the other flies west at a speed of 648 km/hr. How far apart are the airplanes at 5:30 p.m.? Draw a picture and explain.
**Explanation:**
To solve this problem, we need to find the distance between two airplanes moving in perpendicular directions after a certain period. This can be visualized as a right triangle, where each leg of the triangle represents the distance traveled by each airplane, and the hypotenuse represents the distance between the two airplanes.
- **Time Calculation:**
From 2:00 p.m. to 5:30 p.m. is 3.5 hours.
- **Distance Calculation:**
- The airplane flying south travels at 376 km/hr:
\[
\text{Distance South} = 376 \, \text{km/hr} \times 3.5 \, \text{hours} = 1316 \, \text{km}
\]
- The airplane flying west travels at 648 km/hr:
\[
\text{Distance West} = 648 \, \text{km/hr} \times 3.5 \, \text{hours} = 2268 \, \text{km}
\]
- **Using the Pythagorean Theorem:**
With these distances, use the Pythagorean theorem to find the hypotenuse (distance apart):
\[
\text{Distance Apart} = \sqrt{(1316^2 + 2268^2)}
\]
\[
\text{Distance Apart} = \sqrt{(1731856 + 5143824)} = \sqrt{6875680} \approx 2622 \, \text{km}
\]
Thus, the airplanes are approximately 2622 kilometers apart at 5:30 p.m.
**Diagram Explanation:**
- Imagine a right triangle:
- The vertical leg represents the path of the airplane flying south (1316 km).
- The horizontal leg represents the path of the airplane flying west (2268 km).
- The hypotenuse is the direct distance between the airplanes (approximately 2622 km).
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