Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

A wind with velocity 45 miles per hour is blowing in the direction \(\mathrm{N} 20^{\circ} \mathrm{W}\). An airplane that flies at 425 miles per hour in still air is supposed to fly straight north. How should the airplane be headed and how fast will it then be flying with respect to the ground?

Step-by-Step Solution

Answer

The airplane should head approximately N 2.07° E and will fly at about 398.51 mph with respect to the ground.

Step 1: Understand the Problem

The airplane needs to counteract the wind to maintain a straight northward path. The wind's velocity is given, as well as the airplane's speed in still air. We need to determine the heading of the airplane and its effective ground speed.

Step 2: Break Down Wind Velocity

The wind direction is \( ext{N} 20^{\circ} ext{W}\), meaning it blows 20 degrees west of north. Decompose it into components using trigonometry. The northward component is \(-45 \cos(20^{\circ})\) and the westward component is \(-45 \sin(20^{\circ})\).

Step 3: Find Airplane Heading

The airplane needs a heading with eastward components equal and opposite to the wind's westward components to maintain a straight northern path. Let \( heta\) be the angle east of north and solve for it: \(425 \sin(\theta) = 45 \sin(20^{\circ})\). From this, \(\theta = \arcsin\left(\frac{45 \sin(20^{\circ})}{425}\right).\)

Step 4: Calculate Effective Ground Speed

Now, calculate the effective northward ground speed by considering both the airplane's and wind's northward components: \(425 \cos(\theta) - 45 \cos(20^{\circ})\).

Step 5: Solve for Values

Calculate step-by-step: The westward wind component is \(45 \sin(20^{\circ}) \approx 15.34\) mph. Use this in \(\theta = \arcsin\left(\frac{15.34}{425}\right)\approx 2.07^{\circ}\). The northward ground speed is \(425 \cos(2.07^{\circ}) - 45 \cos(20^{\circ}) = 425 (0.999) - 45 (0.9397) \approx 398.51 \,\text{mph}\).

Key Concepts

Wind Velocity

When analyzing the motion of an object, such as an airplane, in an environment with wind, it's crucial to understand wind velocity. The wind velocity affects how the object navigates through the air. Wind velocity is a vector quantity, which means it has both magnitude and direction. In our problem, the wind is blowing at 45 miles per hour in a direction of North 20° West.

This means the wind isn't blowing directly north but is 20 degrees off towards the west. To analyze this, we can decompose the wind velocity into two components using trigonometry:

• The northward component: This can be calculated using the cosine of the angle, i.e., \(-45 \cos(20^{\circ})\).
• The westward component: This is found using the sine, i.e., \(-45 \sin(20^{\circ})\).

By understanding these components, we can begin to understand how they interact with and affect the airplane's trajectory.

Airplane Heading

In still air, an airplane can travel straight in the direction it is aimed. However, when wind is present, its heading needs to adjust to account for the wind's influence, so as to reach the intended direction. In this exercise, the airplane wants to fly straight north despite the wind blowing towards the northwest.

To counteract the westward component of the wind, the airplane will need to head slightly east of north. The component of the airplane's velocity must be equal and opposite to the wind’s westward force. We set up the equation \(425 \sin(\theta) = 45 \sin(20^{\circ})\) to solve for the angle \(\theta\), which determines how much east of north the airplane should head.

By rearranging and calculating, we find that:\( \theta = \arcsin\left(\frac{45 \sin(20^{\circ})}{425}\right)\) This lets us find the correct heading needed to maintain a northward path.

Ground Speed Calculation

Once the airplane's heading is adjusted to counteract the wind, it's important to calculate the effective ground speed. Ground speed is the speed at which the airplane moves relative to the ground, considering both its own speed and the effect of the wind.

To find the effective northward ground speed:

• Calculate the northward component of the airplane's velocity using the corrected heading: \(425 \cos(\theta)\).
• Subtract the wind's northward influence: \(- 45 \cos(20^{\circ})\).

Finally, putting this together gives us the northward ground speed, which is:\(425 \, (0.999) - 45 \, (0.9397) \approx 398.51 \text{ mph} \).This calculation shows how even with wind, the effective speed and direction can be accurately determined.