Problem 53 - Precalculus

Question

A plane flew \(\mathrm{N} 30^{\circ} \mathrm{W}\) at 350 mph for 2.5 hours. \(\mathrm{A}\) second plane, starting at the same point and at the same time, flew \(35^{\circ}\) at an angle clockwise from due north at 550 mph for 2.5 hours. At the end of 2.5 hours, how far apart were the two planes? Round to the nearest mile.

Step-by-Step Solution

Answer

The planes are approximately 1279 miles apart.

Step 1: Calculate the Distance of Each Plane

First, we need to find the distance each plane traveled. **Plane 1** flew at 350 mph for 2.5 hours, so its distance is \(350 \text{ mph} \times 2.5 \text{ hours} = 875 \text{ miles}\). **Plane 2** flew at 550 mph for 2.5 hours, so its distance is \(550 \text{ mph} \times 2.5 \text{ hours} = 1375 \text{ miles}\).

Step 2: Break Down the Vector Components of Plane 1

**Plane 1** flies \(30^\circ\) west of due north. To find its components:- Northward component: \( 875 \times \cos(30^\circ) = 757.77 \text{ miles} \)- Westward component: \( 875 \times \sin(30^\circ) = 437.5 \text{ miles} \)

Step 3: Break Down the Vector Components of Plane 2

**Plane 2** flies at an angle of \(35^\circ\) clockwise from due north, effectively flying \(35^\circ\) east of due north. To find its components:- Northward component: \( 1375 \times \cos(35^\circ) = 1127.52 \text{ miles} \)- Eastward component: \( 1375 \times \sin(35^\circ) = 787.92 \text{ miles} \)

Step 4: Calculate the Total Displacement Between the Planes

To find how far apart they are, consider the difference in their northward and east-west components. - The northward distance apart: \( 1127.52 - 757.77 = 369.75 \text{ miles} \)- The east-west distance apart: \( 787.92 + 437.5 = 1225.42 \text{ miles} \) (since one is westward and the other eastward)Use the Pythagorean theorem: \[ \text{Distance} = \sqrt{(369.75)^2 + (1225.42)^2} \approx \sqrt{136715.06 + 1501195.58} \approx \sqrt{1637910.64} \approx 1279 \text{ miles} \]

Step 5: Round the Answer to the Nearest Mile

The calculated distance between the two planes after 2.5 hours is approximately 1279 miles. Round this to the nearest mile, which is **1279 miles**.

Key Concepts

Vector Components

When dealing with the movement of objects in different directions, understanding vector components is crucial. Vectors represent both magnitude and direction, which is essential in navigation and physics. For instance, when a plane flies at an angle, its path can be broken down into horizontal and vertical components. This is where trigonometry comes into play. By using trigonometric functions like sine and cosine, we can determine how far an object moves in each specific direction.

Suppose a plane flies north-west or east of north; it doesn't simply go entirely north or entirely in another direction. Instead, its actual path can be represented as two movements: one towards the north and another towards the west or east. For example, using the angles provided, we calculate the northward (vertical) and westward or eastward (horizontal) components of each plane's travel using:

• Northward component: Multiply the distance by the cosine of the angle.
• Eastward or Westward component: Multiply the distance by the sine of the angle.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in mathematics but especially significant when dealing with vector components and distances. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Applied in our scenario, once we know the horizontal and vertical separations between two points – which in this case are the vector components of the distance each plane traveled – we can find the direct distance between them (like drawing a line connecting the two) by using the Pythagorean theorem:\[\text{Distance} = \sqrt{(\text{northward component difference})^2 + (\text{east-west component difference})^2}\]

This is how we determine how far apart the two planes are by the end of their journeys.

Distance Calculation

To calculate how far an object or person has traveled, multiply speed by time. For each plane in our example, the basic formula is:

• Distance = Speed × Time

This calculation is straightforward and results in the entire distance each plane covered in 2.5 hours.

After determining each plane’s travel distance, we further used their speeds and directions to break this overall distance into vector components (discussed earlier) for additional analysis. The accurate computation of distance using both the fundamental distance formula and the Pythagorean theorem is critical in real-world navigation and tracking scenarios.

Angle Measurement

Understanding angle measurements is vital in navigation, especially when it involves angles other than the typical cardinal directions (North, South, East, West). Directions are often expressed using angles measured from one of these cardinal points, usually north. For example:

• A plane flying \(35^{\circ}\) clockwise from due north indicates its heading is slightly eastward of north.
• Another at \(30^{\circ}\) westward of north highlights its northwest directionality.

These angle measurements assist in determining the specific paths objects like planes must follow, and aid in calculating vector components. Properly interpreting these angles and applying them in calculations ensures the accuracy of direction and distance estimations on navigation tasks.