A particles moves in a straight line with the acceleration shown in the motion curve given. Knowing that the it starts from the origin with initial velocity of -18 ft per second, plot the velocity - time and distance time curves within 0 to 20 seconds. Compute also the velocity, position and total distance traveled at t = 12 seconds.
A particles moves in a straight line with the acceleration shown in the motion curve given. Knowing that the it starts from the origin with initial velocity of -18 ft per second, plot the velocity - time and distance time curves within 0 to 20 seconds. Compute also the velocity, position and total distance traveled at t = 12 seconds. Show COMPLETE STEP-BY-STEP SOLUTIONS and SHOW NECESSARY FIGURES. Thank you
![a (ft/s²)
6
3
0
-5
4
10
t(s)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49dded77-738c-41b3-aa8d-1c82a87effea%2F4ad9f123-a3ba-49c1-a4ed-173991f73faa%2F3whea1_processed.png&w=3840&q=75)
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Given Question: A particles moves in a straight line with the acceleration shown in the motion curve given. Knowing that the it starts from the origin with initial velocity of -18 ft per second, plot the velocity - time and distance time curves within 0 to 20 seconds. Compute also the velocity, position and total distance traveled at t = 12 seconds.
Solution: https://www.bartleby.com/questions-and-answers/a-particles-moves-in-a-straight-line-with-the-acceleration-shown-in-the-motion-curve-given.-knowing-/4ad9f123-a3ba-49c1-a4ed-173991f73faa
Follow-up Question on solution: How did they get 160m, 736m, and 1488m? Thank you
![a (ft/s²)
6
3
0
-5
4
10
t(s)](https://content.bartleby.com/qna-images/question/49dded77-738c-41b3-aa8d-1c82a87effea/dd9c21ba-8103-4a3a-9978-511818408f0b/mv19ho3_thumbnail.png)
![at t = 11sec
at t= 12sec
att=
-= aosec
velocity
*+/57/
42
18
5
-73
Origin at om
V = -18-5X11 = -731t/sec
V-18-5X20 =~11871/see
A
V=-18-5X20 =
10
P= total Distance
11
Distance time curve
8=0
duplacement = s = ut +/bat²
-118f1/5e0
20 t/sec
at t=0
8 => -18X0+ 1/29X0 = om
2
at t=asec s= -1SX4 +16₂ X3x4² = - 48m
2
at t=10sec s→ 48x10+1/XEX 10 = 120m
al t= 205C S=> +8x20+1/X(-5)x20³⇒-1360m
at t=17SEC S>>-18x12+1/2x5x12²-2576m
at t = 6 sec
F
om
48m
160m
Displacement becomez0o- in between asecdiosec
-let +/6x6x+²=0
31²-1st =0 give t=6sec
S=0
~14.88m
736m](https://content.bartleby.com/qna-images/question/49dded77-738c-41b3-aa8d-1c82a87effea/dd9c21ba-8103-4a3a-9978-511818408f0b/o4c29kuj_thumbnail.jpeg)
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