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EXAMPLE6 A particle moves along a line so that its velocity at time t is v(t) = t-t- 20 (measured in meters per second). (a) Find the displacement of the particle during 2 sts 8. (b) Find the distance traveled during this time period. SOLUTION (a) By this equation, the displacement is 8. s(8) – s(2) = v(t) dt (t2 - t- 20) dt %3D t2 20t 2 This means that the particle moved approximately 18.00 meters to the right. (b) Note that v(t) = t - t - 20 = (t – 5)(t + 4) and so v(t)? 0 on the interval [2, 5] and v(t) ? o on [5, 8]. Thus, from this equation, the distance traveled is |v(t)| dt = [-v(t)] dt + v(t) dt I (--2 + t + 20) dt + (2 - t- 20) dt %3D 18

EXAMPLE6 A particle moves along a line so that its velocity at time t is v(t) = t-t- 20 (measured in meters per second). (a) Find the displacement of the particle during 2 sts 8. (b) Find the distance traveled during this time period. SOLUTION (a) By this equation, the displacement is 8. s(8) – s(2) = v(t) dt (t2 - t- 20) dt %3D t2 20t 2 This means that the particle moved approximately 18.00 meters to the right. (b) Note that v(t) = t - t - 20 = (t – 5)(t + 4) and so v(t)? 0 on the interval [2, 5] and v(t) ? o on [5, 8]. Thus, from this equation, the distance traveled is |v(t)| dt = [-v(t)] dt + v(t) dt I (--2 + t + 20) dt + (2 - t- 20) dt %3D 18

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section: Chapter Questions
Problem 18T
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EXAMPLE 6 A particle moves along a line so that its velocity at time t is v(t) =t-t- 20 (measured in meters per second). (a) Find the displacement of the particle during 2 sts 8. (b) Find the distance traveled during this time period. SOLUTION (a) By this equation, the displacement is 8 s(8) – s(2) = v(t) dt | (2 -t- 20) dt %3D 18 t2 - 20t %3D This means that the particle moved approximately 18.00 meters to the right. (b) Note that v(t) = t - t - 20 = (t - 5)(t + 4) and so v(t) ? 0 on the interval [2, 5] and v(t) ? o on [5, 8]. Thus, from this equation, the distance traveled is |v(t)| dt = [-v(t)] dt + v(t) dt (-t + t + 20) dt + dt 18 ||
Transcribed Image Text:EXAMPLE 6 A particle moves along a line so that its velocity at time t is v(t) =t-t- 20 (measured in meters per second). (a) Find the displacement of the particle during 2 sts 8. (b) Find the distance traveled during this time period. SOLUTION (a) By this equation, the displacement is 8 s(8) – s(2) = v(t) dt | (2 -t- 20) dt %3D 18 t2 - 20t %3D This means that the particle moved approximately 18.00 meters to the right. (b) Note that v(t) = t - t - 20 = (t - 5)(t + 4) and so v(t) ? 0 on the interval [2, 5] and v(t) ? o on [5, 8]. Thus, from this equation, the distance traveled is |v(t)| dt = [-v(t)] dt + v(t) dt (-t + t + 20) dt + dt 18 ||
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