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Math 202 Project
Diving Deception Pass
The problem
• This problem is concerned with the behavior of water as its flowing through the Deception Pass which is a narrow channel between two islands near Seattle, Washington.
The problem is divided into
6 questions. Each one of those questions
represents a solution of a certain part of
the main problem.
Problem 1
Suppose that the initial velocity v(0) of the current under bridge is 7 mph. Convert it to ft/s and find the velocity of the current 600 ft west of the bridge. Likewise find the velocity of the current 200 ft east of the bridge, and 800 ft east of the bridge.
The Equation needed for this part (1)
200000
)100(1)(
)('
2
xxS
xS
Vox
Converting the velocity form mph to ft/s:
7 mph=10.2667 ft/sNow we will use equation 1 :For x=600 to the west V=2.98 ft/sFor x=200 ft to the eastV=9.78 ft/sFor x=800 ft to the westV=2.98 ft/s
Problem 2
Under the assumptions of Problem 1, where is the
current fastest?
The analysis of the data shows that the point where the maximum velocity of the current will occur at x=100 ft which is located east of the bridge.
X (ft)X’)ft/s)
10002.03
7503.30
5005.70
10010.27
09.78
-1008.56
-5003.67
-7502.23
-10001.46
Problem 3
Use the deception pass tool on the de tools cd to see how your motion
changes depending on where you start. In words describe your motion if you enter the channel 1000 feet to the west of the bridge and compare
it to the motion if you start 1000 feet east of the bridge.
The DE Tools CD contains a visual representation of the
motion of the diver during the entire trip. In this part of
the problem we will use two concepts to distinguish between the two cases. The first is equation 1 and
the second is the fact that the maximum velocity of the
current will occur at the narrowest point of the channel.
For the first case were the diver dives 1000 feet west of the bridge he will be moving with an initial velocity that well
keep increasing as the channel gets narrower in tell he reaches the
maximum velocity 100 feet east of the bridge which is the narrowest part of
the channel. After that the velocity will start decreasing as the channel gets wider in tell the end of the motion.
In contrary if the diver starts his motion 1000 feet east of the bridge he will start with an initial velocity that well decrease in tell the end of the motion due to the fact that the channel is getting wider. The difference
between the two cases is that in case 1 the diver will swim through the point of
maximum velocity which is 100 feet east of the bridge unlike case 2 where the diver will
start after that point and in this case the velocity will keep going down.
Velocity during the trip
x vs x'
0
2
4
6
8
1 0
1 2
-1500 -1000 -500 0 500 1000 1500
Problem 4
Suppose that the time t=0 corresponds to slack water, compute the velocity of the current near your
entry point 1000 feet west of the bridge, that is, with x=-1000,when
you started your dive one hour after slack water .remember that time is
given in seconds.
Equation 3
)(
)23400/sin('
xS
tx
In order to determine the velocity of the current 1000 feet west of the bridge and
one hour after the slack water we will have to use equation 2 which is the correction of equation 1 but we need to convert the time
into seconds:Time= 1 hour 3600 s
X=-1000So x' = v0 sin (π *3600/23400) / (1+(-1000-
100)^2/200000)=0.89 ft/s
Problem 5
You want to compute your likely position relative to the bridge from the time you entered the water (1
hour after the slack water and 1000 feet west of the bridge). What initial value problem would you solve to do
so.
This part of the problem needs the use of the initial value concept so we will have
to settle on the parameters of our unknowns that we need to find and
what information do we have in order to design the required initial value
problem. Essentially we will have to use equation 2 as our main differential
equation but we need to determine the initial value condition.
What do we have in our disposal of information are the initial position and the time at which we want to find the required position. Since equation 2
gives the velocity in terms of the time then we have the value of one of the needed parameters which is the time however what we need is a condition that contains both time and position.
This can be explained by the fact that the integral of the velocity is the
position which we have so the needed condition will be x (3600) = -1000
And the needed equation in this case to be solved along with this condition is
equation (3) in the book
Problem 6
Show that the model given in the book (4) gives opposite signs for the
velocity of the current at x = -300 and x = 300 . What does this
mean physically discuss the physical validity of this differential equation in view of the sign of the channel-size
function S(x).
The correction model for the velocity equation
(Equation 3)
1000300
6003)(
)23400/sin('
1000300
6003)(
xxxS
tVox
xxxS
Equation 3 well be used to calculate the corrected velocity
at the following cases:
1) X = - 300
2) X= 300
Using equation 3:
1)x'= 2.73 ft/s
2)x'= - 20.9 ft/s
Here both velocities are measured with using the correctional model (5) and this results
confirms the range of validity of this correctional model which is from x= -1200
to x=0
Case1
in the first case the positive result of the velocity at x= -300 confirm the fact that at that point the correctional model S (x) is valid and yields the expected positive
velocity
S(x)2 Vs x
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
-600 -400 -200 0 200 400 600
Case 2
however in this case it pretty clear that the result obtained for the velocity confirms
that the point x= 300 is beyond the range of the correctional model of S (x)
x'2 Vs x
-25
-20
-15
-10
-5
0
5
10
15
-600 -400 -200 0 200 400 600
The legitimacy of these results agrees with the sings of channel-size function which
gives positive sign for the first case and a negative sign for the second case which lie beyond the range of validity of S (x) which
varies from x=-1200 to x =0 and form here we conclude the physical meaning of this variation of values which clearly show that the range of the correctional model
was exceeded
Q & ADone by:Al-Bassam, A.MAl-Bassam, R.AAl-Dukair, A.M