Question

A migrating robin flies due north with a speed of \(12 \mathrm{~m} / \mathrm{s}\) relative to the air. The air moves due east with a speed of \(6.1 \mathrm{~m} / \mathrm{s}\) relative to the ground. What is the robin's speed relative to the ground?

Short answer

The robin's speed relative to the ground is approximately 13.46 m/s.

Step-by-step solution

Understand the vectors involved

We have two vectors to consider: vector \( \mathbf{v_{RA}} \) of the robin relative to air, with magnitude 12 m/s heading north, and vector \( \mathbf{v_{AG}} \) of the air relative to the ground, with magnitude 6.1 m/s heading east.

Identify direction and components

The robin's velocity vector \( \mathbf{v_{RA}} \) is vertically directed northward, represented as \( \langle 0, 12 \rangle \). The air's velocity vector \( \mathbf{v_{AG}} \) is horizontally directed eastward, represented as \( \langle 6.1, 0 \rangle \).

Use vector addition to find resultant velocity

The robin's velocity relative to the ground \( \mathbf{v_{RG}} \) is the vector sum of bird's velocity relative to the air and the air's velocity relative to the ground: \( \mathbf{v_{RG}} = \mathbf{v_{RA}} + \mathbf{v_{AG}} = \langle 0, 12 \rangle + \langle 6.1, 0 \rangle = \langle 6.1, 12 \rangle \).

Calculate the magnitude of the resultant vector

The speed of the robin relative to the ground can be found by calculating the magnitude of the resultant vector \( \mathbf{v_{RG}} \). This is done using the Pythagorean theorem: \( |\mathbf{v_{RG}}| = \sqrt{6.1^2 + 12^2} \).

Perform calculations

Compute the magnitude: \( |\mathbf{v_{RG}}| = \sqrt{6.1^2 + 12^2} = \sqrt{37.21 + 144} = \sqrt{181.21} \approx 13.46 \mathrm{~m/s} \).

Key concepts

Velocity Vectors

Velocity vectors are crucial in understanding motion. They help define both speed and direction, indicating how different objects or entities move in a given space. In our exercise, we deal with two main velocity vectors:

• The robin's velocity relative to the air, moving north at 12 m/s.
• The air's velocity relative to the ground, moving east at 6.1 m/s.

Velocity vectors differ from just scalar speeds, as vectors incorporate directions, making them a more comprehensive description of motion.
Therefore, when analyzing the robin's movement, understanding its velocity's vector component is essential. By examining these vectors as components on a coordinate system, it becomes easier to visualize and calculate the resultant motion.

Pythagorean Theorem

The Pythagorean theorem is a mathematical concept used here to find the magnitude of the resultant vector. This theorem states: for any right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the two other sides.
In this exercise, we have two vectors forming a right angle:

• The northward velocity of the robin relative to the air, represented vertically as 12 m/s.
• The eastward velocity of the air relative to the ground, represented horizontally as 6.1 m/s.

Using these as the perpendicular components of a right triangle, the hypotenuse is our desired resultant velocity. Applying the Pythagorean theorem: \[|\mathbf{v_{RG}}| = \sqrt{(6.1)^2 + (12)^2}\]This approach helps simplify the calculation of the resultant speed, by treating it as a geometric problem involving right triangles.

Resultant Velocity

Resultant velocity is the combination of multiple velocity vectors to describe an object's overall movement relative to a specific frame of reference. In our case, the robin's trajectory relative to the ground.The key steps to determine resultant velocity through vector addition are:

• Identify each velocity vector and its direction.
• Add these vectors component-wise.
• Calculate the resultant's magnitude using the Pythagorean theorem.

For our exercise, the robin’s resultant velocity relative to the ground is found as follows:

• Sum the robin’s and air vectors to get \( \mathbf{v_{RG}} = \langle 6.1, 12 \rangle \).
• Calculate the magnitude \( \sqrt{6.1^2 + 12^2} \approx 13.46 \text{ m/s} \).

Thus illustrating the complete description of motion, and allowing the understanding of complex movements by breaking them into simpler, componential analysis.