Problem 37 - University Physics with Modern Physics

Question

Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. If one goose is flying at 100 km/h relative to the air but a 40-km/h wind is blowing from west to east, (a) at what angle relative to the north-south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of 500 km from north to south? (\(Note\): Even on cloudy nights, many birds can navigate by using the earth's magnetic field to fix the north-south direction.)

Step-by-Step Solution

Answer

(a) The goose should head 66.42° west of south; (b) it will take approximately 5.46 hours.

Step 1: Determine the Velocity of the Wind

The wind is blowing from west to east at 40 km/h. Therefore, the wind's velocity vector is purely in the eastward direction.

Step 2: Calculate the Required Southward Velocity

The goose wants to travel in a straight north-south line directly southward. So, we need the resultant vector of the bird's velocity (which includes the wind's effect) to have no east-west component and only a southward component. Therefore, we need to counteract the 40 km/h eastward wind component.

Step 3: Sketch the Right Triangle

Imagine a right triangle where:- The hypotenuse is the goose's velocity relative to the air (100 km/h).- The southward leg (vertical component) is the velocity the goose needs to maintain directly south relative to the ground.- The eastward leg (horizontal component) is the wind speed (40 km/h).Let's denote the angle the goose must head relative to the north-south direction as \( \theta \).

Step 4: Apply Trigonometric Functions

To find the angle \( \theta \), we use the cosine function since \( \cos(\theta) = \frac{40}{100} \).By solving, we get:\[ \theta = \cos^{-1} \left( \frac{40}{100} \right) = \cos^{-1}(0.4) \approx 66.42^\circ \].Thus, the goose must head at an angle of approximately \( 66.42^\circ \) west of south.

Step 5: Determine Southward Component of the Goose's Velocity

Using the Pythagorean theorem: the southward velocity can be computed as \( v_y = \sqrt{(100^2 - 40^2)} \).Solving this, we have: \( v_y = \sqrt{10000 - 1600} = \sqrt{8400} \approx 91.65 \text{ km/h} \).

Step 6: Calculate Travel Time

The ground distance to be covered southward is 500 km. With the effective southward speed being approximately 91.65 km/h, the time \( t \) can be calculated by:\[ t = \frac{500}{91.65} \approx 5.46 \text{ hours} \].

Key Concepts

Vector Addition

In physics, vectors are a way to represent quantities that have both magnitude and direction. When dealing with vectors, such as wind and velocity, understanding how to add them correctly is important. In our exercise with the goose and the wind, we deal with vector addition to determine the effective path the goose needs to fly.

In this problem, we use the concept of vector addition to find how the goose's velocity combines with the wind velocity. The bird flies at 100 km/h relative to the air, not necessarily directly south. To fly directly southward against a 40 km/h wind from the west, the goose must aim slightly southwest. This creates a right triangle, with the hypotenuse representing the bird’s velocity in the air, the eastward component as the wind, and the desired southward component.

Vector addition involves solving how these components interact, using trigonometry, to achieve a resultant vector that tackles both the goose's direction and the wind's influence. This step is crucial to find the correct angle and speed to mitigate the wind’s effect.

Angle Calculation

Calculating the angle at which the goose should be flying involves trigonometry. This helps determine the right head adjustment to counteract the eastward push of the wind. We need an angle that ensures the goose moves directly south despite the wind.

By considering the velocity vectors involved, we use the cosine trigonometric function. Since the cosine function relates the adjacent side (wind speed) to the hypotenuse (bird's velocity), we have:

• Wind speed (adjacent) = 40 km/h
• Goose's airspeed (hypotenuse) = 100 km/h

The cosine of the angle is given by the ratio of these sides, \[ \cos(\theta) = \frac{40}{100} = 0.4 \]Solving for the angle \( \theta \) allows us to find:\[ \theta = \cos^{-1}(0.4) \approx 66.42^\circ \]
This calculation shows that the goose must head approximately \( 66.42^\circ \) west of true south to keep flying directly southward relative to the ground.

Velocity Components

Understanding velocity components is essential in breaking down the goose's velocity into usable parts. These components include the southward velocity (parallel to the direction the goose intends to travel) and the eastward velocity (caused by wind interference).

We begin by using the Pythagorean theorem to solve for the southward component of the goose's velocity:\[ v_y = \sqrt{(100^2 - 40^2)} \]
This gives:\[ v_y = \sqrt{10000 - 1600} = \sqrt{8400} \approx 91.65 \text{ km/h} \]
This result tells us that the effective velocity of the goose moving southward relative to the earth is approximately 91.65 km/h.

At this calculated velocity, covering the ground distance of 500 km south will take:\[ t = \frac{500}{91.65} \approx 5.46 \text{ hours} \]
Knowing these components helps us predict real-world outcomes and effectively plan the goose’s migration against external influences like wind.