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
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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Question
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e273367-affb-43ef-886e-3231d3c073b4%2Fef82df91-28ba-4056-84a3-e6da2508b23f%2Fm6orme_processed.jpeg&w=3840&q=75)
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
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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