Collin Chegg.com Wiley PLUS C edugen.wileyplus.com/edugen/student/mainfr.uni Ass

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Collin Chegg.com Wiley PLUS C edugen.wileyplus.com/edugen/student/mainfr.uni Assignment Open Assignment ONEXTD FULL SCREEN PRINTER VERSION HBACK ASSIGNMENT RESOURCES Chapter 04, Problem 020 Homework 3 Fall 2015 Chapter 04 Concept In the figure, particle A moves along the line y 28 m with a constant velocity v of magnitude 2.5 m/s and directed parallel to the x axis. At the instant particle A passes the y axis Ouestion 02 Chapter 04 Problem 001 particle B leaves the origin with zero initial speed and constant acceleration a of magnitude 0.48 m/s2. What angle e between a and the positive direction of the y axis would Chapter 04 Problem 002 Chapter 04 Problem result in a collision? 003 Chapter 04 Problem 005 Chapter 04 Problem 006 SN Chapter 04 Problem 007 Chapter 04 Problem 008 Chapter 04 Problem Chapter 04. Problem 013 SN Chapter 04 Problem Chapter 04 Problem Number Units Chapter 04 Problem the tolerance is +/-2% 020 License Agreement I Privacy Policy I 2000-2015 John Wiley Sons Inc. All Rights Reserved. A Division of John Wiley & Sons Inc. Version 4.16.1.7 ss 9/10/2015 6:50 PM

Explanation / Answer

let c be the poiint where ray meet and let B has to be equal to that distance travelled by Ae

so

x = 28 tan theta

time taken = ta = 28 tan thtea/2.5 ----------------------1

for same time, B will reach C,

so for B, S = ut + 0.5 at^2

28/cos theta = 0.24t^2 -----------------------2

from 1 and 2

28/cos theta = 0.24 * 28 * 28 * tan theta^2/(2.5*2.5)

cos theta * tan^2 theta = 0.939

1- cos^2 Theta = 0.939 costheta

cos^2 theta + 0.939 cos theta -1 = 0

so solving for cos theta

cos theta = 0.635 or 1.6 this is not possible

so

theta = 50.58 deg   

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