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.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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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)
Transcribed Image Text:a (ft/s²) 6 3 0 -5 4 10 t(s)
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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)
Transcribed Image Text:a (ft/s²) 6 3 0 -5 4 10 t(s)
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
Transcribed Image Text: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
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