Question

A plane leaves the airport in Galisteo and flies 170 km at 68.0\(^{\circ}\) east of north; then it changes direction to fly 230 km at 36.0\(^{\circ}\) south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?

Short answer

Fly 351.5 km at 11.7° south of east.

Step-by-step solution

Resolve First Leg of the Journey

The plane first travels 170 km at an angle of 68.0 degrees east of north. We need to find the northern and eastern components of this leg of the journey. The northern component can be calculated using the cosine function: \[ 170 \times \cos(68.0^\circ) \approx 63.9 \text{ km} \] The eastern component can be calculated using the sine function: \[ 170 \times \sin(68.0^\circ) \approx 158.1 \text{ km} \]

Resolve Second Leg of the Journey

Next, the plane travels 230 km at an angle of 36.0 degrees south of east. We need to calculate the south and east components of this leg. The eastern component is calculated using cosine:\[ 230 \times \cos(36.0^\circ) \approx 186.1 \text{ km} \] The southern component is calculated using sine: \[ 230 \times \sin(36.0^\circ) \approx 135.2 \text{ km} \]

Calculate Net Displacement Components

Now, we combine the vectors from steps 1 and 2 to find the net displacement components. The total northern displacement is:\[ 63.9 \text{ km (north) } - 135.2 \text{ km (south) } = -71.3 \text{ km} \] (The result is negative indicating the net direction is south.)The total eastern displacement is:\[ 158.1 \text{ km} + 186.1 \text{ km} = 344.2 \text{ km} \]

Find Total Displacement and Direction

Now, determine the magnitude of the total displacement using the Pythagorean theorem:\[ \sqrt{(344.2)^2 + (71.3)^2} \approx 351.5 \text{ km} \]To find the direction, calculate the angle from the east using the tangent function:\[ \theta = \tan^{-1}\left(\frac{71.3}{344.2}\right) \approx 11.7^\circ \] This angle is south of east.

Key concepts

Understanding Displacement Components

When we talk about displacement components, we're essentially breaking down a vector into parts that make it easier to analyze. Think about how a plane's journey can be split into 'north'/'south' and 'east'/'west' components.

Each movement of the plane can be seen as a vector, which has both magnitude (distance) and direction (angle). When a plane travels at an angle, like north-east or south-east, we don't directly know how far it goes in the north or east direction separately. Instead, we resolve the vector into its components.

• Northern Component: How far the plane travels northward.
• Eastern Component: How far the plane travels eastward.
• Southern Component: How far the plane travels southward, if reversing direction.

We use trigonometric functions to find these components. For angles measured from the north, cosine gives us the northern component and sine gives us the eastern component. Conversely, for angles measured from the east, cosine gives us the eastern component and sine gives us the southern or northern component, depending on the direction.

Applying the Pythagorean Theorem

The Pythagorean theorem is a fundamental principle used to find the relation between the sides of a right triangle. It states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

In the context of vector addition for displacement, once we have the net components in each direction, we can use the Pythagorean theorem to find the total displacement. Here’s how:

• Net Eastern and Northern/Southern Components: Once we have resolved all vectors into their components, we'll have a net eastern and a net northern or southern component.
• Calculate Hypotenuse (Total Displacement): Apply the formula \( d = \sqrt{(x)^2 + (y)^2} \) where \( x \) and \( y \) are the net east and south/north components.

By calculating this, you'll get the straight-line distance the rescue team needs to fly directly to the plane's location.

Role of Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are essential when working with angles and vectors. They help us resolve a vector's magnitude into its horizontal and vertical components or find angles between vectors. Here’s a quick refresher on how they work:

• Sine (\(\sin\)): In a right triangle, it is the ratio of the length of the opposite side to the hypotenuse; useful for finding vertical components from angles.
• Cosine (\(\cos\)): This is the ratio of the length of the adjacent side to the hypotenuse; used for horizontal components from angles.
• Tangent (\(\tan\)): The ratio of the opposite side to the adjacent side; ideal for finding an angle when you know two sides.

For example, when a plane travels at a 68-degree angle east of north, \( \cos(68.0^\circ) \) helps find the northern component, and \( \sin(68.0^\circ) \) finds the eastern component. To determine direction, \( \tan^{-1} \left( \frac{opposite}{adjacent} \right) \) helps us find the angle from net components, expressing the direction from an axis.