2. The displacement of a particle which moves along the x axis is given by x = (-4 + 31²)e-0.251, consider x to be in feet and t in seconds. Plot the displace- ment, velocity and acceleration for the first 20 seconds of motion and determine, both graphically and by your established equation for acceleration, the time at which acceleration is 0. (If the students find it more comfortable, they may choose to graph using excel or any graphing software, solutions however must be done manually)
2. The displacement of a particle which moves along the x axis is given by x = (-4 + 31²)e-0.251, consider x to be in feet and t in seconds. Plot the displace- ment, velocity and acceleration for the first 20 seconds of motion and determine, both graphically and by your established equation for acceleration, the time at which acceleration is 0. (If the students find it more comfortable, they may choose to graph using excel or any graphing software, solutions however must be done manually)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Can you please make it look like this? Also the plot. Please Thanks!
![GIVEN: X (-4+3t)e-0.3t
SOLUTION:
v: ds
dt
Sds Svdt
ved
² [(-4+3+) e ²0:3]
dt
#
V(e-03) (3) + (-4+3+) (e-0.3+) (0.3)
y = 3e -0.3 (3t-4)e-0.3t
V. e 0.3t
(1.8-1.5t)
a. dv
dt
a:
d
dt
a e
-0.3t
a e
[e-0.3 (1.8-1.5 t)]
-0-8t
a e-0.3t [1.5+ (1.8-1.5 +) (-0.3)]
-0.3t
(-1.5-0.54 + 0.75+)
(0.75t - 2.04)
-0.3t
a.eo.
(1.5) + (1.8-1.5t) (e-0.²) (-0.3)
NOW, DETERMINE ACCELERATION WHICH IS 0.
a 0 t?
0=e
t. 2.04
7.5
t 2 72s
-0.3t
[0.75 -2.04]](https://content.bartleby.com/qna-images/question/3284259c-f7b4-4f03-a9ad-5639bb0f9627/53a00c2c-fdcb-4b39-8cf6-6172300d6bdd/gu1tuzb_processed.jpeg)
Transcribed Image Text:GIVEN: X (-4+3t)e-0.3t
SOLUTION:
v: ds
dt
Sds Svdt
ved
² [(-4+3+) e ²0:3]
dt
#
V(e-03) (3) + (-4+3+) (e-0.3+) (0.3)
y = 3e -0.3 (3t-4)e-0.3t
V. e 0.3t
(1.8-1.5t)
a. dv
dt
a:
d
dt
a e
-0.3t
a e
[e-0.3 (1.8-1.5 t)]
-0-8t
a e-0.3t [1.5+ (1.8-1.5 +) (-0.3)]
-0.3t
(-1.5-0.54 + 0.75+)
(0.75t - 2.04)
-0.3t
a.eo.
(1.5) + (1.8-1.5t) (e-0.²) (-0.3)
NOW, DETERMINE ACCELERATION WHICH IS 0.
a 0 t?
0=e
t. 2.04
7.5
t 2 72s
-0.3t
[0.75 -2.04]
Transcribed Image Text:PLOT
DISPLACEMENT
-2
VELOCITY TIME
4
3
2
=
2
-3
TIME
ACCELERATION TIME
>
Solution
by Bartleby ExpertKnowledge Booster
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