/*! 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 8 Find the \(x\) and \(y\) compone... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the \(x\) and \(y\) components of a position vector \(\overrightarrow{\mathrm{r}}\) of magnitude \(r=75 \mathrm{m},\) if its angle relative to the \(x\) axis is (a) \(35.0^{\circ}\) and (b) \(65.0^{\circ}\).

Short Answer

Expert verified
For \( 35^{\circ} \), \( r_x = 61.44 \text{ m} \), \( r_y = 43.02 \text{ m} \); for \( 65^{\circ} \), \( r_x = 31.70 \text{ m} \), \( r_y = 67.97 \text{ m} \).

Step by step solution

01

Understand the Problem

The position vector \( \overrightarrow{\mathrm{r}} \) has a magnitude of \( r = 75 \text{ m} \) and we need to find its \( x \) and \( y \) components based on given angles relative to the \( x \)-axis.
02

Breakdown the Vector Components

The \( x \) component of the vector, \( r_x \), is calculated using \( r_x = r \cos \theta \). The \( y \) component, \( r_y \), is calculated using \( r_y = r \sin \theta \), where \( \theta \) is the angle relative to the \( x \)-axis.
03

Calculate Components for Angle 35°

Given \( \theta = 35.0^{\circ} \), compute:\[ r_x = 75 \cos(35^{\circ}) \]\[ r_y = 75 \sin(35^{\circ}) \]Using a calculator, we find:- \( r_x \approx 75 \times 0.8192 = 61.44 \text{ m} \)- \( r_y \approx 75 \times 0.5736 = 43.02 \text{ m} \)
04

Calculate Components for Angle 65°

Given \( \theta = 65.0^{\circ} \), compute:\[ r_x = 75 \cos(65^{\circ}) \]\[ r_y = 75 \sin(65^{\circ}) \]Using a calculator, we find:- \( r_x \approx 75 \times 0.4226 = 31.695 \text{ m} \)- \( r_y \approx 75 \times 0.9063 = 67.9725 \text{ m} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Position Vector
A position vector is a vector that represents the position or location of a point in space relative to an origin. It is often denoted as \( \overrightarrow{r} \) and typically characterized by its magnitude and direction. When given a position vector, understanding its components is crucial. The vector itself is oriented in a particular direction, and its components express how far it extends in each coordinate direction. In a two-dimensional plane, the position vector can be broken down into horizontal \( (x) \) and vertical \( (y) \) components. By analyzing the position vector in this way, we can easily determine the exact location of the point it represents.In practical scenarios, you'll often be given the magnitude of the vector and an angle relative to one of the axes, like the \( x \)-axis. This is where understanding the relation between the angle and the components comes into play, allowing for precise calculations and representations.
Trigonometric Functions
Trigonometric functions, specifically sine and cosine, play a pivotal role in determining the components of any vector given its angle with respect to an axis. These functions relate the angles in a right triangle to the ratios of its sides, making them ideal for splitting a vector into its components.For any angle \( \theta \) with the positive \( x \)-axis:
  • The cosine of the angle, \( \cos \theta \), provides the ratio of the adjacent side \( (x \text{ component}) \) to the hypotenuse \( (magnitude) \).
  • The sine of the angle, \( \sin \theta \), reflects the ratio of the opposite side \( (y \text{ component}) \) to the hypotenuse.
Given this information, the \( x \) and \( y \) components of a vector can be easily computed:
  • \( r_x = r \cos \theta \)
  • \( r_y = r \sin \theta \)
By plugging in the values of the vector magnitude and the angle, we can derive the numerical values of these components, which are essential in both physics and engineering applications.
Vector Magnitude
The magnitude of a vector is a measure of its length or size. When dealing with a position vector, it refers to the straight-line distance from the origin to the vector's endpoint. The magnitude is a scalar quantity, meaning it only has size and no direction.In our specific exercise, the magnitude \( r \) represents how far the point is from the origin along a straight path. With a value of \( 75 \text{ m} \), it shows the vector's length without any directional information yet.When decomposing the vector into its \( x \) and \( y \) components, the magnitude remains constant, but its directional attributes are revealed through the components. These components squared and summed (using the Pythagorean theorem) will always equal the square of the magnitude, ensuring that the vector is accurately represented as:
  • \( r = \sqrt{r_x^2 + r_y^2} \)
Understanding vector magnitude is essential in accurately representing vectors in any dimensional space and is crucial for calculations involving force, velocity, and more.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two people take identical Jet Skis across a river, traveling at the same speed relative to the water. Jet Ski A heads directly across the river and is carried downstream by the current before reaching the opposite shore. Jet Ski B travels in a direction that is \(35^{\circ}\) upstream and arrives at the opposite shore directly across from the starting point. (a) Which Jet Ski reaches the opposite shore in the least amount of time? (b) Confirm your answer to part (a) by finding the ratio of the time it takes for the two Jet Skis to cross the river. (Note: Angles are measured relative to the \(x\) axis shown in Example 3-2.)

The Velocity of the Moon The velocity of the Moon relative to the center of the Earth can be approximated by $$\begin{aligned} \overrightarrow{\mathbf{v}}=&(945 \mathrm{m} / \mathrm{s})\left\\{-\sin \left[\left(2.46 \times 10^{-6} \mathrm{radians} / \mathrm{s}\right) t\right] \hat{\mathrm{x}}\right.\\\ &\left.+\cos \left[\left(2.46 \times 10^{-6} \mathrm{radians} / \mathrm{s}\right) t\right] \hat{\mathrm{y}}\right\\} \end{aligned}$$ where \(t\) is measured in seconds. To approximate the instantaneous acceleration of the Moon at \(t=0,\) calculate the magnitude and direction of the average acceleration between the times (a) \(t=0\) and \(t=0.100\) days and (b) \(t=0\) and \(t=0.0100\) days. (The time required for the Moon to complete one orbit is 29.5 days.)

The Longitude Problem In \(1755,\) John Harrison \((1693-1776)\) completed his fourth precision chronometer, the \(\mathrm{H} 4\) which eventually won the celebrated Longitude Prize. (For the human drama behind the Longitude Prize, see Longitude, by Dava Sobel.) When the minute hand of the H4 indicated 10 minutes past the hour, it extended \(3.0 \mathrm{cm}\) in the horizontal direction. (a) How long was the \(\mathrm{H} 4\) 's minute hand? (b) At 10 minutes past the hour, was the extension of the minute hand in the vertical direction more than, less than, or equal to \(3.0 \mathrm{cm}\) ? Explain. (c) Calculate the vertical extension of the minute hand at 10 minutes past the hour.

Two students perform an experiment with a train and a ball. Michelle rides on a flatcar pulled at \(8.35 \mathrm{m} / \mathrm{s}\) by a train on a straight, horizontal track; Gary stands at rest on the ground near the tracks. When Michelle throws the ball with an initial angle of \(65.0^{\circ}\) above the horizontal, from her point of view, Gary sees the ball rise straight up and back down above a fixed point on the ground. (a) Did Michelle throw the ball toward the front of the train or toward the rear of the train? Explain. (b) What was the initial speed of Michelle's throw? (c) What was the initial speed of the ball as seen by Gary?

Vector A points in the negative \(y\) direction and has a magnitude of 5 units. Vector \(\overrightarrow{\mathrm{B}}\) has twice the magnitude and points in the positive \(\underline{x}\) direction. Find the direction and magnitude of (a) \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) (b) \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}},\) and (c) \(\overrightarrow{\mathrm{B}}-\overrightarrow{\mathbf{A}}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.