syms t 60 sin( 161) e x-60"sin(16*t)*exp(-((24*t)/5))% Displacement v=diff(x)% Velocity 960 cos(161) e - 288 sin( 16t) e a-diff(x,2) % Acceleration clear all t-0; -9216 cos( 161) e 69888 sin( 161) e display('displacemt at t-e') x-60*sin(16*t)*exp(-((24*t)/5)) displacemt at t-e display('Velocity at t-0') v=960" cos (16*t)*exp(-((24*t)/5)) 288*sin(16*t)*exp(-((24*t)/5)) Velocity at t=0 V- 960 display('Acceleration at t-0') a=- 9216* cos(16*t)*exp(-((24*t)/5)) - (69888*sin(16*t)*exp(-((24*t)/5)))/5 % Displacement Acceleration at t-0 a- -9216 clear all t-0.3 t- 0.3000 displacemt at t-e x - -14.1611 display('displacemt at t-e') x-60"sin(16*t)*exp(-((24*t)/5)) bisplay('Velocity at t-0.3') v=960*cos (16*t)*exp(-((24*t)/5)) - 288*sin(16*t)*exp(-((24*t)/5)) Velocity at t=0.3 8752. 87 = צ display('Acceleration at t-e.3') a-- 9216*cos(16*t)*exp(-((24*t)/5)) - (69888*sin(16*t)*exp(-((24*t)/5)))/5 % Displacement Acceleration at t-0.3 a- 3.1079e+03
syms t 60 sin( 161) e x-60"sin(16*t)*exp(-((24*t)/5))% Displacement v=diff(x)% Velocity 960 cos(161) e - 288 sin( 16t) e a-diff(x,2) % Acceleration clear all t-0; -9216 cos( 161) e 69888 sin( 161) e display('displacemt at t-e') x-60*sin(16*t)*exp(-((24*t)/5)) displacemt at t-e display('Velocity at t-0') v=960" cos (16*t)*exp(-((24*t)/5)) 288*sin(16*t)*exp(-((24*t)/5)) Velocity at t=0 V- 960 display('Acceleration at t-0') a=- 9216* cos(16*t)*exp(-((24*t)/5)) - (69888*sin(16*t)*exp(-((24*t)/5)))/5 % Displacement Acceleration at t-0 a- -9216 clear all t-0.3 t- 0.3000 displacemt at t-e x - -14.1611 display('displacemt at t-e') x-60"sin(16*t)*exp(-((24*t)/5)) bisplay('Velocity at t-0.3') v=960*cos (16*t)*exp(-((24*t)/5)) - 288*sin(16*t)*exp(-((24*t)/5)) Velocity at t=0.3 8752. 87 = צ display('Acceleration at t-e.3') a-- 9216*cos(16*t)*exp(-((24*t)/5)) - (69888*sin(16*t)*exp(-((24*t)/5)))/5 % Displacement Acceleration at t-0.3 a- 3.1079e+03
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Formulation of the solution and Interpretation of the solution of this. Essay type. Thanks.
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X =
syms t
-241
60 sin( 16) e
2
3
x=60*sin(16*t)*exp(-((24*t)/5))% Displacement
V =
4
V=diff(x)% Velocity
-241
960 cos(161) e5 - 288 sin(161) e
6
a-diff(x,2) % Acceleration
a =
8
241
-24
-9216 cos(16 t) e
69888 sin(16 1) e
9
clear all
10
t=0;
11
display('displacemt at t-0')
displacemt at t-e
12
x-60*sin(16*t)*exp(-((24*t)/5))
13
display('Velocity at t=0')
v=960*cos (16*t)*exp(-((24*t)/5)) - 288*sin(16*t)*exp(-((24*t)/5))
14
Velocity at t=0
15
v = 960
16
display('Acceleration at t=0')
a=- 9216*cos(16*t)*exp(-((24*t)/5)) - (69888*sin(16*t)*exp(-((24*t)/5)))/5 % Displacement
clear all
17
Acceleration at t=0
18
a - -9216
-
19
20
t=0.3
t = 0.3000
display('displacemt at t-e')
x-60*sin(16*t)*exp(-((24*t)/5))
þisplay('Velocity at t=0.3')
v=960*cos (16*t)*exp(-((24*t)/5)) - 288*sin(16*t)*exp(-((24*t)/5))
21
displacemt at t-e
22
x = -14.1611
23
Velocity at t=0.3
24
v = 87.8752
25
26
display('Acceleration at t-0.3')
Acceleration at t-0.3
a=- 9216*cos(16*t)*exp(-((24*t)/5)) - (69888*sin(16*t)*exp(-(( 24*t)/5)))/5 % Displacement
a = 3.1079e+03
27
script
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