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Example 6 Video Example4) A particle moves along a line so that its velocity at time t is v(t) = 2 -t- 12 (measured in meters per second). (a) Find the displacement of the partidle during 2 st 9. (b) Find the distance traveled during this time period. Solution (a) By this equation, the displacement is as follows. s(9) - s(2) = v(t) dt (t2 -t- 12) dt 66 This means that the particle moved approximately 117.83 meters to the right. (b) Note that v(t) = t -t- 12 (t- 4)(t + 3) and so v(t) 2 0 on the interval [2, 4] and v(t) ?0 on (4, 9]. Thus, from the equation I(e)l dt = total distance traveled, the distance traveled is as follows. dt = (2 -t- 12) dt Need Help? Read It

Example 6 Video Example4) A particle moves along a line so that its velocity at time t is v(t) = 2 -t- 12 (measured in meters per second). (a) Find the displacement of the partidle during 2 st 9. (b) Find the distance traveled during this time period. Solution (a) By this equation, the displacement is as follows. s(9) - s(2) = v(t) dt (t2 -t- 12) dt 66 This means that the particle moved approximately 117.83 meters to the right. (b) Note that v(t) = t -t- 12 (t- 4)(t + 3) and so v(t) 2 0 on the interval [2, 4] and v(t) ?0 on (4, 9]. Thus, from the equation I(e)l dt = total distance traveled, the distance traveled is as follows. dt = (2 -t- 12) dt Need Help? Read It

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Example 6 Video Example) A particle moves along a line so that its velocity at time t is v(t) = t2 -t- 12 (measured in meters per second). (a) Find the displacement of the particle during 2 st 9. (b) Find the distance traveled during this time period. Solution (a) By this equation, the displacement is as follows. s(9) - s(2) = v(t) dt (t2 -t- 12) dt 66 This means that the particle moved approximately 117.83 meters to the right. (b) Note that v(t) = t? -t- 12 = (t - 4)(t + 3) and so v(t) ? 0 on the interval [2, 4] and v(t) ?0 on [4, 9]. Thus, from the equation I(t)l dt = total distance traveled, the distance traveled is as follows. dt = (t2 -t- 12) dt Need Help? Read It
Transcribed Image Text:Example 6 Video Example) A particle moves along a line so that its velocity at time t is v(t) = t2 -t- 12 (measured in meters per second). (a) Find the displacement of the particle during 2 st 9. (b) Find the distance traveled during this time period. Solution (a) By this equation, the displacement is as follows. s(9) - s(2) = v(t) dt (t2 -t- 12) dt 66 This means that the particle moved approximately 117.83 meters to the right. (b) Note that v(t) = t? -t- 12 = (t - 4)(t + 3) and so v(t) ? 0 on the interval [2, 4] and v(t) ?0 on [4, 9]. Thus, from the equation I(t)l dt = total distance traveled, the distance traveled is as follows. dt = (t2 -t- 12) dt Need Help? Read It
Video Example (2x - 6x +) dx and interpret the result in terms of areas. Find Solution The Fundamental Theorem gives 6x + 3 +4 tan -극2)-3(22) + 4 tan-(2)-(| -3(0)2 +2 tan-lo |-4 1.- +4 tan (2). This is the exact value of the integral. If a decimal approximation is desired, wwe can use a calculator to approximate tan(2). Doing so, we get 1 7857 (Round your answer to four decimal places.) The figure below shows the graph of the integrand. We know that the value of the integral can be interpreted as a net area: the sum of the areas labeled with a plus sign minus the area labeled with a minus sign.
Transcribed Image Text:Video Example (2x - 6x +) dx and interpret the result in terms of areas. Find Solution The Fundamental Theorem gives 6x + 3 +4 tan -극2)-3(22) + 4 tan-(2)-(| -3(0)2 +2 tan-lo |-4 1.- +4 tan (2). This is the exact value of the integral. If a decimal approximation is desired, wwe can use a calculator to approximate tan(2). Doing so, we get 1 7857 (Round your answer to four decimal places.) The figure below shows the graph of the integrand. We know that the value of the integral can be interpreted as a net area: the sum of the areas labeled with a plus sign minus the area labeled with a minus sign.
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