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second). (a) Find the displacement of the particle during 3 sts 9. (b) Find the distance traveled during this time period. SOLUTION (a) By this equation, the displacement is 6. s(9) – s(3) - vce) dt 20) dt This means that the particle moved approximately 78.00 meters to the right. (b) Note that v(t) = t² - t – 20 = (t - 5)(t + 4) and so v(t) ?v0 on the interval [3, 5] and v(t) ?0 on [5, 9]. Thus, from this equation, the distance traveled is | IvcE)l dt = [-v(t)] dt + t + 20) dt + - 20) dt 19 +

second). (a) Find the displacement of the particle during 3 sts 9. (b) Find the distance traveled during this time period. SOLUTION (a) By this equation, the displacement is 6. s(9) – s(3) - vce) dt 20) dt This means that the particle moved approximately 78.00 meters to the right. (b) Note that v(t) = t² - t – 20 = (t - 5)(t + 4) and so v(t) ?v0 on the interval [3, 5] and v(t) ?0 on [5, 9]. Thus, from this equation, the distance traveled is | IvcE)l dt = [-v(t)] dt + t + 20) dt + - 20) dt 19 +

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter2: Graphical And Tabular Analysis
Section2.4: Solving Nonlinear Equations
Problem 17E: Van der Waals Equation In Exercise 18 at the end of Section 2.3, we discussed the ideal gas law,...
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EXAMPLE 6 A particle moves along a line so that its velocity at time t is v(t) = 2 - t - 20 (measured in meters per second). (a) Find the displacement of the particle during 3 stS 9. (b) Find the distance traveled during this time period. SOLUTION (a) By this equation, the displacement is s(9) – s(3) - v v(t) dt = - (2 -t - 20) dt - [ This means that the particle moved approximately 78.00 meters to the right. (b) Note that v(t) = t² - t - 20 = (t - 5)(t + 4) and so v(t) ?0 on the interval [3, 5] and v(t) ?v0 on [5, 9]. Thus, from this equation, the distance traveled is 6- 5 6. |v(t)| dt = [-v(t)] dt + v(t) dt (-t +t + 20) dt + (t2 - t - 20) dt 15 19 3 a in +
Transcribed Image Text:EXAMPLE 6 A particle moves along a line so that its velocity at time t is v(t) = 2 - t - 20 (measured in meters per second). (a) Find the displacement of the particle during 3 stS 9. (b) Find the distance traveled during this time period. SOLUTION (a) By this equation, the displacement is s(9) – s(3) - v v(t) dt = - (2 -t - 20) dt - [ This means that the particle moved approximately 78.00 meters to the right. (b) Note that v(t) = t² - t - 20 = (t - 5)(t + 4) and so v(t) ?0 on the interval [3, 5] and v(t) ?v0 on [5, 9]. Thus, from this equation, the distance traveled is 6- 5 6. |v(t)| dt = [-v(t)] dt + v(t) dt (-t +t + 20) dt + (t2 - t - 20) dt 15 19 3 a in +
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