Question

The air-speed indicator of a plane that took off from Detroit reads \(350 . \mathrm{km} / \mathrm{h}\) and the compass indicates that it is heading due east to Boston. A steady wind is blowing due north at \(40.0 \mathrm{~km} / \mathrm{h}\). Calculate the velocity of the plane with reference to the ground. If the pilot wishes to fly directly to Boston (due east) what must the compass read?

Short answer

Answer: The ground velocity of the plane is 353.6 km/h. The pilot should fly at an angle of 83.66° from north to reach Boston directly.

Step-by-step solution

Identify the known parameters

The air-speed indicator of the plane reads 350 km/h due east and the wind is blowing due north at 40.0 km/h.

Set up the velocity vectors

We'll use the following notation: - \(V_{PA}\): Velocity of the plane concerning the air (air-speed indicator) - \(V_W\): Velocity of the wind concerning the ground - \(V_{PG}\): Velocity of the plane concerning the ground (which we want to find) We can write the vectors as follows: \(V_{PA} = 350 \ \text{km/h} \ \hat{i}\) \(V_W = 40 \ \text{km/h} \ \hat{j}\)

Calculate the ground velocity of the plane

Using vector addition, we can find the ground velocity of the plane. \(V_{PG} = V_{PA} + V_W\) Plugging in the values of the vectors, we have: \(V_{PG} = (350 \ \text{km/h} \ \hat{i}) + (40 \ \text{km/h} \ \hat{j})\)

Find the magnitude of the ground velocity

Calculate the magnitude of the ground velocity vector using the Pythagorean theorem: \(|V_{PG}| = \sqrt{(V_{PG_x})^2 + (V_{PG_y})^2} = \sqrt{(350 \ \text{km/h})^2 + (40 \ \text{km/h})^2} = 353.6 \ \text{km/h}\) So, the ground velocity of the plane is \(353.6 \ \text{km/h}\) with an upward angle.

Find the direction to fly directly to Boston

In order to fly directly to Boston (due east), the pilot needs to change the direction of the plane. We will find the angle \(\theta\) between the \(y\)-axis (due north) and the ground velocity vector \(V_{PG}\). Using the tangent function, we have: \(\tan(\theta) = \frac{V_{PG_x}}{V_{PG_y}} = \frac{350}{40}\) Now we find the angle: \(\theta = \arctan(\frac{350}{40}) = 83.66^\circ\) Therefore, the pilot needs to change the compass to \(83.66^\circ\) from north.

Key concepts

Vector Addition

The concept of vector addition is fundamental in understanding the physical world around us, particularly when dealing with multiple forces or velocities acting at the same time. In physics, a vector represents both a magnitude and a direction. To find the result of two or more vectors acting concurrently, we use vector addition.

For example, imagine you're pushing a cart eastwards while a friend pushes it northwards. The actual path of the cart will be a combination of both directions. Vector addition allows us to calculate this combined effect mathematically. We often illustrate vectors as arrows on a graph, with their length showing the magnitude and their angle indicating direction. To add vectors, we place them head to tail and draw a new vector (the resultant) from the tail of the first to the head of the second.

In the context of our plane exercise, vector addition helps us compute the plane's ground velocity by combining its airspeed velocity vector with the wind's velocity vector.

Ground Velocity

Ground velocity in physics is the speed and direction with which an object moves across the Earth's surface. It's the actual velocity that would be measured by an observer on the ground. For an aircraft, it's different from its airspeed, which is the speed relative to the air around it.

Ground velocity is important for navigation and is influenced by factors like wind. To calculate it, we take into account the airspeed of the aircraft and the velocity of the wind. Using vector addition, these velocities are combined to find the true path and speed of the aircraft across the ground. The ground velocity is particularly relevant when plotting a course for a destination, as it determines where you will end up, not just where you are pointed.

Air-Speed Indicator

An air-speed indicator is an instrument found on aircraft that measures the speed of the aircraft through the air. It's crucial for pilots as it helps them understand the aircraft's performance and ensures safe flight conditions. However, the reading from an air-speed indicator doesn't reflect the ground velocity; it only tells us how fast the plane is moving relative to the air around it.

This can be misleading if not properly interpreted, especially when there is wind. If a plane is flying with a substantial tailwind, for example, its ground speed will be higher than the airspeed indicates, and if there's a headwind, the ground speed will be lower. Pilots must account for this discrepancy to accurately judge their time of arrival at the destination and ensure they are on the correct flight path.

Wind Velocity

Wind velocity is the speed and direction of the wind, which is a critical factor in the context of flight and other meteorological concerns. For pilots, it is essential to consider wind velocity to maintain the desired course and ensure an efficient flight path. Wind velocity is often a vector quantity, and as such, it is composed of a speed component and a direction.

In our example, wind blowing due north at 40 km/h affects the plane's eastward journey, pushing it off-course. Pilots must adjust their course to compensate for this wind, which is precisely what vector addition helps to determine. In severe weather conditions, understanding wind velocity can be the difference between a smooth flight and a potentially hazardous situation.