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Chapter 4: Problem 74

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

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

Step by step solution

01

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.
02

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 \( \langle 6.1, 0 \rangle \).
03

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: \( \mathbf{v_{RG}} = \mathbf{v_{RA}} + \mathbf{v_{AG}} = \langle 0, 12 \rangle + \langle 6.1, 0 \rangle = \langle 6.1, 12 \rangle \).
04

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: \( |\mathbf{v_{RG}}| = \sqrt{6.1^2 + 12^2} \).
05

Perform calculations

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 resultant velocity relative to the ground is found as follows:
  • Sum the robin鈥檚 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.

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Most popular questions from this chapter

The hang time of a punt is measured to be \(4.50 \mathrm{~s}\). If the ball was kicked at an angle of \(63.0^{\circ}\) above the horizontal and was caught at the same level from which it was kicked, what was its initial speed?

Suppose that each component of a vector is doubled. (a) Does the magnitude of the vector increase, decrease, or stay the same? Explain. (b) Does the direction angle of the vector increase, decrease, or stay the same?

Tower An air traffic controller observes two airplanes approaching the airport. The displacement from the control tower to plane 1 is given by the vector \(\overrightarrow{\mathbf{A}}\), which has a magnitude of \(220 \mathrm{~km}\) and points in a direction \(32^{\circ}\) north of west. The displacement from the control tower to plane 2 is given by the vector \(\overrightarrow{\mathbf{B}}\), which has a magnitude of \(140 \mathrm{~km}\) and points \(65^{\circ}\) east of north. (a) Sketch the vectors \(\overrightarrow{\mathbf{A}},-\overrightarrow{\mathbf{B}}\), and \(\overrightarrow{\mathbf{D}}=\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\). Notice that \(\vec{D}\) is the displacement from plane 2 to plane 1 . (b) Use components to find the magnitude and the direction of the vector \(\overrightarrow{\mathbf{D}}\).

Playing shortstop, you pick up a ground ball and throw it to second base. The ball is thrown horizontally, with a speed of \(22 \mathrm{~m} / \mathrm{s}\), directly toward point \(A\) as shown in Figure 4.37. When the ball reaches the second baseman \(0.45 \mathrm{~s}\) later, it is caught at point B. (a) How far were you from the second baseman? (b) What is the distance of vertical drop, from \(A\) to \(B\) ?

Given that \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{C}}\) and that \(A^{2}+B^{2}=C^{2}\), how are \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) oriented relative to one another?
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