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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "Gabe Rumley" }}{PARA 0 "" 
0 "" {TEXT -1 17 "Number 21 page 96" }}{PARA 0 "" 0 "" {TEXT -1 7 "2-2
3-00" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 465 "
The problem has a sailboat going in a straight line on a body of water
. There is a lght wind pushing it at 1m/sec. Then the wind picks up an
d blows hard enough to apply a constant force of 600 newtons. Water re
sistance is the only other force acting on the boat and it has a propo
rtionality constant of k=100 kg/sec. This water resistance is proporti
onal to the boat's velocity. The sailboat has a mass of 50 kg. I am to
 find the equation of motion of the sailboat." }}{PARA 0 "" 0 "" 
{TEXT -1 125 "So first I found the equation for the velocity shown bel
ow. I used the 1m/sec for my intial velocity as given in the problem.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "v:=unapply(rhs(dsolve(\{diff(v(t),t
)=(600-(100*v(t)))/50,v(0)=1\},v(t))),t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"vGR6#%\"tG6\"6$%)operatorG%&arrowGF(,&\"\"'\"\"\"-%
$expG6#,$9$!\"#!\"&F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 352 "Now
 I use the velocity equation to find the equation of motion. The equat
ion of motion is the antiderivative of the velocity equation. The prob
lem gave no initial condition for the distance, so theorectically you \+
could put the origin any where you wanted. I talk about distance here \+
because that is what x(t) represents; distance as a function of time.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "x:=unapply(rhs(dsolve(
\{diff(x(t),t)=v(t)\},x(t))),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"xGR6#%\"tG6\"6$%)operatorG%&arrowGF(,(9$\"\"'-%$expG6#,$F-!\"##\"\"&
\"\"#%$_C1G\"\"\"F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 574 "Above
 is the equation of motion and as you can see there is a constant C1 i
n the equation; this is becasue there was no initial condition on the \+
distance.Somehow even though there was no intial condition given in th
e book, the answer in the back of the book has C1 defined as 5/2. I do
 not know where they got the initial condition to calculate C1.So I we
nt on without it. Now, I am suppose to find the limiting velocity of t
he sailboat under this wind. This means what is the velocity as time g
oes toward infinty. I plugged in for the time and let maple evaluate i
t for me." }{TEXT 256 81 "Dr. Sarah's Comment - if x(0)=0, then C1 is \+
-5/2 because of the following below:\012" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "_C1:=solve(x(0)=0,_C
1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$_C1G#!\"&\"\"#" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "evalf(v(infinity));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"\"'\"\"!" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 154 "The limit
ing velocity comes out to be 6m/sec. This is because in the velocity e
quation as time goes to infinity that term goes to zero, so 6m/sec is \+
left." }}}}{MARK "0 1 0" 17 }{VIEWOPTS 1 1 0 1 1 1803 }