/*! 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} Q2Q Figure 4-22 shows the initial po... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Figure 4-22 shows the initial position i and the final position f of a particle. What are the (a) initial position vectorr1→and (b) final position vectorrf→, both in unit-vector notation? (c) What is the x component of displacement∆r→?

Short Answer

Expert verified

a) Initial position vectorri→=7iÁåœ+jÁåœ-2kÁåœ

b) Final position vector rf→=5iÁåœ-3jÁåœ+kÁåœ

c) The x component of displacement∆r→=-2iÁåœ

Step by step solution

01

Given information

rix=7 mriy=1 mriz=-2 mrfx=5 mrfy=-3 mrfz=1 m

02

To understand the concept

Vector can be written by measuring the distances along each direction. Displacement can be calculated by change in position between two vectors.

Formulae:

Unit vector notation

r→=rxiÁåœ+ryjÁåœ+rzkÁåœ

03

(a) To find initial position vector r1→

Initial position vector r1→

r1→=rixiÁåœ+riyjÁåœ+rizkÁåœr1→=7iÁåœ+jÁåœ-2kÁåœ

04

(b) To find initial position vector rf→

rf→=rfxiÁåœ+rfyjÁåœ+rfzkÁåœrf→=5iÁåœ-3jÁåœ+kÁåœ

05

(c) To find x component of displacement ∆r→

We have, displacement vector is given by:

∆r→=rf→-ri→=5-7iÁåœ+-3-1jÁåœ+1--2kÁåœ=-2iÁåœ-4jÁåœ+3kÁåœ

Hence,

The x component of the displacement is ∆r→=-2iÁåœ.

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!

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

Snow is falling vertically at a constant speed of 8.0m/s. At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of 50km/h?

You are to throw a ball with a speed of 12.0m/sat a target that is height
h=5.00mabove the level at which you release the ball (Fig. 4-58). You want the ball’s velocity to be horizontal at the instant it reaches the target. (a) At what angle above the horizontal must you throw the ball? (b) What is the horizontal distance from the release point to the target? (c) What is the speed of the ball just as it reaches the target?

An airport terminal has a moving sidewalk to speed passengers through a long corridor. Larry does not use the moving sidewalk; he takes 150s to walk through the corridor. Curly, who simply stands on the moving sidewalk, covers the same distance in 70s. Moe boards the sidewalk and walks along it. How long does Moe take to move through the corridor? Assume that Larry and Moe walk at the same speed.

In Fig. 4-32, particleAmoves along the linelocalid="1654157716964" y=30mwith a constant velocitylocalid="1654157724548" v⇶Äof magnitude localid="1654157733335" 3.0m/sand parallel to thelocalid="1654157740474" xaxis. At the instant particlelocalid="1654157754808" Apasses thelocalid="1654157747127" yaxis, particlelocalid="1654157760143" Bleaves the origin with a zero initial speed and a constant accelerationlocalid="1654157772700" aof magnitudelocalid="1654157766903" 0.40m/s2. What anglelocalid="1654157780936" θbetween aand the positive direction of thelocalid="1654157787877" yaxis would result in a collision?

A particle moves so that its position (in meters) as a function of time (in seconds) is r⇶Ä=i^+4t2j^+tk^. Write expressions for: (a) its velocity and (b) its acceleration as functions of time.

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.