A particle starts at rest and moves in a straight line with acceleration a =any constants from integration.)The velocity of the particle as a function of time is v(t) =The velocity of the particle at t = 9 seconds is v(9)The position of the particle at time t is s(t) =coefficients.The position of the particle at time t = 9 is s(9) =-2t+1 m/s². (Note: Omitm/sm/sm. Include 3 significant digits in themAdd WorkSubmit Questionesc90000000F1F2Ma

Given, a=-2t+1 m/s^2Velocity of particle as a function of time:we know…

Answered: A particle starts at rest and moves in…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A particle is moving in three dimensions and its position vector is given by; r→(t)=(1.6t^2+3.7t)i+(3.7t−1.4)j+(4.6t^3+3.8t)k where r is in meters and ?t is in seconds. Determine the magnitude of the average acceleration between t=0s and t=2s. Express your answer in units of m/s2 using one decimal place.
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If a particle’s position is given by the expression x (t) =3.4(t²− 2x −8)³ − 5.4(t²− 2x −80)²+12) meters, what is the velocity acceleration of the particle after t = 5 seconds?
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A particle moving along a straight line has initial velocity v(0) = 1 m/s, and acceleration a(t) = sin(I-t) + cos(t+1) (in m/s?). (a. Find the velocity at time t. (b) Find the average velocity during the time interval [I, 211]. (c) Find the distance traveled during the time interval [1, 2I].
The velocity of a particle traveling in a straight line is given by v = (6t – 3t²) m/s, where t is in seconds. If s = 0 when t = 0, determine the particle's deceleration and position when t = 5 s. How far has the particle traveled during the t = 5 s time interval, and what is its average speed? I need a clear answer by hand, not by keyboard | dybala

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