1. [FTC6] Assume that a particle moves along a straight line and that its velocity is given by the function v(t) = 3t22t 1 in meters per second where time t is given by -1 ≤ t≤ 1. Use the graph of v(t) below to answer the questions. Set up the problem and solve. v(t) 312-21-1 8 7 6 5. 3 2 a. Find the displacement or net distance travelled by the particle in the time interval [−1,1]. Indicate the units. b. Shade the region described by the displacement or net distance. V(t) = 3t² - 2+ -| = 13--13-2 1² - -1 = 0 2-0-1=0 c. Find the distance that the particle travels in the time interval [−1, 1]. Include the units. d. On the graph, shade the region described by the distance. v(t) 312-21-1 8 7 6 5 4 3 2 =64 নि 27 2+4 - v(t) = 3 t² - 2t − 1 = 0 a = 3 b=-2 C = -1 = -(-2) ± √√√2² - 4 (3) (1) 2±√4+12 2(3) t = 2 + 4 = | 6 Velocity changes = = = 1/3 + + = 1 - 3 t 2 = 2-1 - -1/3 3 - ( 3+ ² - 2+ -1) = - 3 +² + 2+ +| = 3 +² - 2+-1 ·S.³ (- 31 + 2 + + 1) dt + S' Distance = -521 27 こ +=-) 1 - ( - 21/12 ) = 1 + (3²-2+ -1d+ = 32 2 2±√16 4 ון = 2nd Intergral t = t=- -૧ 27 + -32 27 દર્ " S 6 of 7 3|~ 64 27
1. [FTC6] Assume that a particle moves along a straight line and that its velocity is given by the function v(t) = 3t22t 1 in meters per second where time t is given by -1 ≤ t≤ 1. Use the graph of v(t) below to answer the questions. Set up the problem and solve. v(t) 312-21-1 8 7 6 5. 3 2 a. Find the displacement or net distance travelled by the particle in the time interval [−1,1]. Indicate the units. b. Shade the region described by the displacement or net distance. V(t) = 3t² - 2+ -| = 13--13-2 1² - -1 = 0 2-0-1=0 c. Find the distance that the particle travels in the time interval [−1, 1]. Include the units. d. On the graph, shade the region described by the distance. v(t) 312-21-1 8 7 6 5 4 3 2 =64 নि 27 2+4 - v(t) = 3 t² - 2t − 1 = 0 a = 3 b=-2 C = -1 = -(-2) ± √√√2² - 4 (3) (1) 2±√4+12 2(3) t = 2 + 4 = | 6 Velocity changes = = = 1/3 + + = 1 - 3 t 2 = 2-1 - -1/3 3 - ( 3+ ² - 2+ -1) = - 3 +² + 2+ +| = 3 +² - 2+-1 ·S.³ (- 31 + 2 + + 1) dt + S' Distance = -521 27 こ +=-) 1 - ( - 21/12 ) = 1 + (3²-2+ -1d+ = 32 2 2±√16 4 ון = 2nd Intergral t = t=- -૧ 27 + -32 27 દર્ " S 6 of 7 3|~ 64 27
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.3: The Natural Exponential Function
Problem 44E
Related questions
Question
Im confused on part d
![1. [FTC6] Assume that a particle moves along a straight line and that its velocity is given by the function
v(t) = 3t22t 1 in meters per second where time t is given by -1 ≤ t≤ 1. Use the graph of v(t) below to
answer the questions. Set up the problem and solve.
v(t) 312-21-1
8
7
6
5.
3
2
a. Find the displacement or net distance travelled by the particle in the time interval [−1,1]. Indicate the units.
b. Shade the region described by the displacement or net distance.
V(t) = 3t² - 2+ -|
=
13--13-2
1² - -1 = 0
2-0-1=0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2a2f0c8-65a9-4348-91e9-ca05b52c3a62%2F3692bc62-17c8-4f29-be5f-eda56613b30b%2Ffdoqae_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. [FTC6] Assume that a particle moves along a straight line and that its velocity is given by the function
v(t) = 3t22t 1 in meters per second where time t is given by -1 ≤ t≤ 1. Use the graph of v(t) below to
answer the questions. Set up the problem and solve.
v(t) 312-21-1
8
7
6
5.
3
2
a. Find the displacement or net distance travelled by the particle in the time interval [−1,1]. Indicate the units.
b. Shade the region described by the displacement or net distance.
V(t) = 3t² - 2+ -|
=
13--13-2
1² - -1 = 0
2-0-1=0
![c. Find the distance that the particle travels in the time interval [−1, 1]. Include the units.
d. On the graph, shade the region described by the distance.
v(t) 312-21-1
8
7
6
5
4
3
2
=64
নि
27
2+4
-
v(t) = 3 t² - 2t − 1 = 0
a = 3
b=-2
C = -1
=
-(-2) ± √√√2² - 4 (3) (1) 2±√4+12
2(3)
t = 2 + 4 = |
6
Velocity changes
= = = 1/3 + + = 1
-
3
t
2 = 2-1 - -1/3
3
- ( 3+ ² - 2+ -1) = - 3 +² + 2+ +|
= 3 +² - 2+-1
·S.³ (- 31 + 2 + + 1) dt + S'
Distance =
-521
27
こ
+=-)
1 - ( - 21/12 ) = 1 +
(3²-2+ -1d+
=
32
2
2±√16
4
ון
=
2nd Intergral
t = t=-
-૧
27 +
-32
27
દર્
"
S
6 of 7
3|~
64
27](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2a2f0c8-65a9-4348-91e9-ca05b52c3a62%2F3692bc62-17c8-4f29-be5f-eda56613b30b%2F1wnyxkp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:c. Find the distance that the particle travels in the time interval [−1, 1]. Include the units.
d. On the graph, shade the region described by the distance.
v(t) 312-21-1
8
7
6
5
4
3
2
=64
নि
27
2+4
-
v(t) = 3 t² - 2t − 1 = 0
a = 3
b=-2
C = -1
=
-(-2) ± √√√2² - 4 (3) (1) 2±√4+12
2(3)
t = 2 + 4 = |
6
Velocity changes
= = = 1/3 + + = 1
-
3
t
2 = 2-1 - -1/3
3
- ( 3+ ² - 2+ -1) = - 3 +² + 2+ +|
= 3 +² - 2+-1
·S.³ (- 31 + 2 + + 1) dt + S'
Distance =
-521
27
こ
+=-)
1 - ( - 21/12 ) = 1 +
(3²-2+ -1d+
=
32
2
2±√16
4
ון
=
2nd Intergral
t = t=-
-૧
27 +
-32
27
દર્
"
S
6 of 7
3|~
64
27
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