uate the position, velocity and acceleration when t = 3 seconds.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:1. The velocity of a particle which moves along the a linear reference axis is given by v = 2—4t-6t³, † is in seconds while v is in meters per second. Eval-
uate the position, velocity and acceleration when t = 3 seconds. Assume an your own initial position and initial point in time. Further, set a variable for
position as you see fit.
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can you please show the integration in the position? And can you please make it look like this? Thanks
![GIVEN: V-2-4t - St
t 3 SECONDS
SOLUTION :
v=2-4t-5t
V= ds
dt
s=vSdt
s. S(2-4t-st³) dt
S=
S-2t -2+²-5 (²3) + ² + c
S = 2t-2+²-2t² + c
S = 2(t-t²- + ³ + c)
dv
dt
a =
a
:
(2-4t-5t 2/2 ) dt
4-5 ( ²2/2) + 2/
-
a = ·4-15 + 2/
REQUIRED: S = ?
V = ?
a = ?
@t3 SECONDS
52 [3-(3) ² (3) ¾ ] +3
S=
- 40.18 m
v=2-[4(3)]-5 (3) ²
v=-35.98 m/s
a =
-4-15/+2
a = - 4 - [ ¹1/21 (3) ¹² ]
a = -16.99 m/s ²
2](https://content.bartleby.com/qna-images/question/3284259c-f7b4-4f03-a9ad-5639bb0f9627/e627a778-75be-4bc7-aaca-a72b698314cd/5z6l0r8_processed.jpeg)
Transcribed Image Text:GIVEN: V-2-4t - St
t 3 SECONDS
SOLUTION :
v=2-4t-5t
V= ds
dt
s=vSdt
s. S(2-4t-st³) dt
S=
S-2t -2+²-5 (²3) + ² + c
S = 2t-2+²-2t² + c
S = 2(t-t²- + ³ + c)
dv
dt
a =
a
:
(2-4t-5t 2/2 ) dt
4-5 ( ²2/2) + 2/
-
a = ·4-15 + 2/
REQUIRED: S = ?
V = ?
a = ?
@t3 SECONDS
52 [3-(3) ² (3) ¾ ] +3
S=
- 40.18 m
v=2-[4(3)]-5 (3) ²
v=-35.98 m/s
a =
-4-15/+2
a = - 4 - [ ¹1/21 (3) ¹² ]
a = -16.99 m/s ²
2
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