/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A jetliner is moving at a speed ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 1

A jetliner is moving at a speed of \(245 \mathrm{~m} / \mathrm{s}\). The vertical component of the plane's velocity is \(40.6 \mathrm{~m} / \mathrm{s}\). Determine the magnitude of the horizontal component of the plane's velocity.

Short Answer

Expert verified
The horizontal component of the plane's velocity is approximately 241.52 m/s.

Step by step solution

01

Understanding the Problem

We are given the total speed of a jetliner, which is its resultant velocity, and its vertical velocity component. To find the horizontal velocity component, we'll use the Pythagorean theorem because the velocities form a right triangle.
02

Applying the Pythagorean Theorem

The Pythagorean theorem states that the sum of the squares of the perpendicular components equals the square of the hypotenuse. In this case, the horizontal velocity component (\( v_h \)) and the vertical velocity component form the right triangle whose hypotenuse is the total speed of the jetliner \( v = 245 \, \text{m/s} \). So, we can write:\[v_h^2 + v_v^2 = v^2\]where \(v_v = 40.6 \, \text{m/s}\) is the vertical component.
03

Solving for the Horizontal Component

Re-arrange the Pythagorean theorem equation to solve for the horizontal component:\[v_h^2 = v^2 - v_v^2\]Substitute the known values to find \(v_h\):\[v_h^2 = (245)^2 - (40.6)^2\]\[v_h^2 = 60025 - 1648.36\]\[v_h^2 = 58376.64\]
04

Calculating the Horizontal Component

Find the square root of \(58376.64\) to determine \(v_h\):\[v_h = \sqrt{58376.64} \approx 241.52 \, \text{m/s}\]

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Key Concepts

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

Pythagorean theorem in physics
The Pythagorean theorem is traditionally associated with right triangles. In physics, it's not just a mathematical footnote; it's crucial for analyzing motion, particularly when vectors are involved. Imagine a right triangle. Its sides can represent different aspects of motion, such as velocity components. The Pythagorean theorem then helps us resolve these components into a single, resultant vector. In the context of the jetliner's motion:
  • One side represents the vertical velocity component.
  • The other side, the horizontal velocity component.
  • The hypotenuse stands for the total speed or the resultant velocity of the jetliner.
The theorem is expressed as:\[ v_h^2 + v_v^2 = v^2 \]This equation allows us to find a missing component as long as the other two are known. It's significant in physics as it simplifies the analysis of complex motions, providing precise answers to otherwise challenging problems.
Velocity components
In physics, velocity is more than just a number; it's a vector, meaning it has both magnitude and direction. Understanding how velocity can be broken down into components is necessary when working with two-dimensional motion.
This is often related to the coordinate system:
  • The horizontal component measures how fast an object moves along the x-axis.
  • The vertical component measures the speed along the y-axis.
For the jetliner example, we know the components form a right triangle, so we apply the Pythagorean theorem. By understanding these components, one can determine how they contribute to the overall velocity, allowing for precise calculations of other quantities, like impact speed, travel time, and more.
Jetliner speed
The speed of a jetliner is a fascinating topic because it presents a rich example of physics in action. Jetliners travel at very high speeds, and understanding how these speeds are determined involves breaking them down into components.
In the given exercise, we were provided the total speed, which is the hypotenuse of our velocity triangle. We were tasked with finding a missing piece: the horizontal component. To understand this in practical terms, consider:
  • Total speed: It's the combination of both vertical and horizontal movements.
  • Horizontal speed: Critical for navigation and determining the flight's course and duration.
  • Vertical speed: Often related to the climb or descent of the plane.
Utilizing the Pythagorean theorem isn't just academic; it's practical, helping pilots and engineers in crafting safe and efficient flight paths. Being able to compute components of speed ensures accurate navigation and fuel efficiency.

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Most popular questions from this chapter

Concept Simulation 3.2 at reviews the concepts that are important in this problem. A golfer imparts a speed of \(30.3 \mathrm{~m} / \mathrm{s}\) to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. (a) How much time does the ball spend in the air? (b) What is the longest "hole in one" that the golfer can make, if the ball does not roll when it hits the green?

Two cars, \(A\) and \(B,\) are traveling in the same direction, although car \(A\) is \(186 \mathrm{~m}\) behind car \(\mathrm{B}\). The speed of \(\mathrm{A}\) is \(24.4 \mathrm{~m} / \mathrm{s},\) and the speed of \(\mathrm{B}\) is \(18.6 \mathrm{~m} / \mathrm{s}\). How much time does it take for A to catch B?

Interactive Solution \(\underline{3.11}\) at presents a model for solving this problem. The earth moves around the sun in a nearly circular orbit of radius \(1.50 \times 10^{11} \mathrm{~m} .\) During the three summer months (an elapsed time of \(7.89 \times 10^{6} \mathrm{~s}\) ), the earth moves one-fourth of the distance around the sun. (a) What is the average speed of the earth? (b) What is the magnitude of the average velocity of the earth during this period?

(a) Does Barbara see him moving toward the east or toward the west? (b) Does Barbara see him moving toward the north or toward the south? (c) Considering your answers to parts (a) and (b), how does Barbara see Neil moving relative to herself, toward the east and north, toward the east and south, toward the west and north, or toward the west and south? Justify your answers in each case. With respect to the ground, Barbara is skating due south at a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). With respect to the ground, Neil is skating due west at a speed of \(3.2 \mathrm{~m} / \mathrm{s} .\) Find Neil's velocity (magnitude and direction relative to due west) as seen by Barbara. Make sure that your answer agrees with your answer to part (c) of the Concept Questions.

An airplane with a speed of \(97.5 \mathrm{~m} / \mathrm{s}\) is climbing upward at an angle of \(50.0^{\circ}\) with respect to the horizontal. When the plane's altitude is \(732 \mathrm{~m}\), the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.
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