Problem 82 - College Physics

Question

An airplane flies due north with an air speed of \(250 \mathrm{~km} / \mathrm{h}\). A steady wind at \(75 \mathrm{~km} / \mathrm{h}\) blows eastward. (Air speed is the speed relative to the air.) (a) What is the plane's ground speed \(\left(v_{\mathrm{pg}}\right) ?\) (b) If the pilot wants to fly due north, what should his heading be?

Step-by-Step Solution

Answer

The plane's ground speed is approximately 261 km/h, and the heading should be approximately 16.7° west of north to fly due north.

Step 1: Understanding Components of Velocity

To determine the plane's ground speed and heading, we need to analyze both the air speed and wind velocity as components of a vector. The plane's air speed is directed completely northward at 250 km/h, and the wind speed is directed completely eastward at 75 km/h. These two velocities form a right triangle.

Step 2: Using the Pythagorean Theorem for Ground Speed

The plane's ground speed takes into account both the northward and eastward velocity components. By treating these as perpendicular components of a triangle, we apply the Pythagorean theorem:\[ v_{pg} = \sqrt{(250)^2 + (75)^2} \]

Step 3: Calculating the Ground Speed

Calculate the values inside the square root:\[ v_{pg} = \sqrt{62500 + 5625} = \sqrt{68125} \approx 261 \text{ km/h} \]

Step 4: Determining the Plane's Heading with Trigonometry

To determine the correct heading so that the plane flies due north relative to the ground, we use trigonometry. Since the wind blows eastward, the plane must be steered slightly west of north to maintain a true north course. We use the inverse tangent function to find the angle:\[ \theta = \tan^{-1} \left( \frac{75}{250} \right) \]

Step 5: Calculating the Heading Angle

Calculate the angle using the inverse tangent function:\[ \theta = \tan^{-1} \left( \frac{75}{250} \right) \approx \tan^{-1} (0.3) \approx 16.7^\circ \] east of north. To counteract this, the plane needs to head approximately 16.7° west of north.

Key Concepts

Velocity Components

When we talk about velocity components in vector analysis, we are breaking down a velocity vector into its parts along perpendicular axes. For example, consider an airplane flying due north at a certain speed. If there is a steady wind blowing from the west to the east, the airplane's actual path, known as ground speed and direction, is influenced by this additional wind component. Imagine the velocity components much like walking in a room: you might be moving towards the door (north), but if there's a breeze pushing you sideways (east), you'll end up going in a diagonal line.

To calculate the effective direction and speed, you need to examine each part separately. For the airplane, the northward flight speed is one component, and the eastward wind speed is the other. These two velocities form the sides of a right triangle, allowing us to understand the total impact on the airplane's movement.

Pythagorean Theorem

The Pythagorean theorem can help us determine the resultant length or magnitude of a vector in a two-dimensional space. It states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. This is expressed mathematically as:\[ c^2 = a^2 + b^2 \]In the context of our airplane problem, the hypotenuse represents the plane's ground speed, while the other two sides represent the northward and eastward velocity components. If the airplane flies due north at 250 km/h, and an eastward wind blows at 75 km/h, we can imagine these as the two short sides of our right triangle. We apply the Pythagorean theorem to find the ground speed:\[ v_{pg} = \sqrt{(250)^2 + (75)^2} \]This calculation combines both components into a single number, giving the effective speed of the plane over the ground.

Trigonometry

Trigonometry is the study of the relationships between the angles and sides of triangles, especially right triangles. In navigation and physics, it's commonly used to ascertain angles and directions.

In our airplane scenario, trigonometry helps determine the angle at which the plane should adjust its heading to counteract the wind. Even though the plane flies northward, the eastward wind causes a deviation, necessitating the use of trigonometric functions such as sine, cosine, and tangent. By knowing two sides of the right triangle (northward and eastward components), you can determine the necessary course adjustment.

Specifically, we use the tangent function, which relates the angle of the triangle to the opposite and adjacent sides. This becomes crucial when determining how much to adjust the plane's direction to maintain a northern trajectory.

Inverse Tangent Function

The inverse tangent function, or arctan, is used to calculate an angle when we know the opposite and adjacent sides of a right triangle. It is a vital tool in vector analysis, navigation, and physics.

For our airplane problem, when we align the velocity components, the inverse tangent function helps determine the adjustment angle from the north. By calculating:\[ \theta = \tan^{-1} \left( \frac{75}{250} \right) \]We find the angle \(\theta\), which represents the deviation caused by the eastward wind speed. Here, \(75\) km/h corresponds to the wind (opposite side) and \(250\) km/h to the plane's velocity in the north direction (adjacent side).

The result is the angle the pilot must adjust to keep the actual flying direction pointed due north on the ground. Understanding this concept ensures accurate navigation in various practical situations.