4%) gnment BACK MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION Chapter 04, Pr

Question

4%) gnment BACK MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION Chapter 04, Problem 015 Your answer is partially correct. Try again. V 7.501 m/s and a constant acceleration A particle leaves the origin with an initial velocity a454-2.31 mis2. When the particle reaches its maximum x coordinate, what are (a) its velocity, (b) its position vector? (a) Number 3.8115 Units m/s (b) Numberi-18.55 TI 3.144 Units m Click if you would like to Show Work for this question: Open Show Work By accessing this Question Assistance, you will learn while you earn points based on the Point Potential instructor Policy set by your Question Attempts: 1 of 4 used SAVE FOR LATER SUBMIT ANSWER

Explanation / Answer

given

initial velocity is

u = 7.5 i m/s

from the kinematic equation

vx = ux + at

at maximum the final velocity of the object is zero vx = 0

0 = ux + ax t

t= -ux/ ax = -7.5/-4.54 = 1.6519 s

vx= 7.5 i + ( -4.54i - 2.31j) (1.6519 s)

= 7.5 i -7.5 i -3.8115 j

= - 3.8115 j

(b)

s= ut + 1/2 at^2

= (7.5 i) 1.6519) + 0.5 ( -4.54 i -2.31 j) ( 1.6519)^2

= 12.38 i -6.194 i -3.15 j

= 6.195 i -3.15 j

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