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R#mX#__#Jc11^^^$$^^^^^^Deception Pass Bridge and the channel beneath it has been known for along while to the people of Seattle, Washington as one of their favorite holiday destinations spots due to the spectacular view of the channel and the great appreciation and welcome that they receive from the locals. Despite the great view of the channel the small village doesnt satisfy the entertainment needs of the young visitors so they try to satisfy those needs by trying taking boat trips or trying a swim once in a while. However, a rather dangerous entertainment method appeared to the surface over the last few years which contains jumping into the channel from a very high point in the adjacent mountain and without a great deal of attention injuries could occur. In this project the differential equation will be used to clarify the safety of those jumps which could save a lot of lives which occur due to this problem .This problem is concerned with the behavior of water as its flowing through the channel.
The problem is divided into 6 questions. Each one of those questions represents a solution of a certain part of the main problem.
x' = v0 / (1+(x100)2/200000) ( (1)
x' = v0 sin( t/23400) / (1+(x100)2/200000) ( (2)
1:
7 mile/hour=10.2667 ft/s
For x=600 to the west
V=2.98 ft/s
For x=200 ft to the east
V=9.78 ft/s
For x=800 ft to the west
V=2.98 ft/s
2:
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.
3:
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 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 he will start after that point and in this case the velocity will keep going down.
4:
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=1000
So x' = v0 sin( *3600/23400) / (1+(1000100)^2/200000)=0.89 ft/s
5:
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 need condition will be X (3600)= 1000
6:
For the last part of the problem the book gives a correction model for the velocity equation:
x' = v0 sin( t/23400) / (3+x+600/300x/1000) ( (3)
Equation 3 well be used to calculate the corrected velocity at the following cases:
x =  300 , t = 3600s , V(o)=13.5 ft/s
in this case x'= 2.73 ft/s
x = 300
in this case x'=  20.9 ft/s
Here both velocities represent a distinguishable physical meaning and in order to explain the sign difference between at the two cases could be clarified by the use of the following analysis:
in this case west of the bridge the direction diagram shows:
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.
in this case east of the bridge the direction diagram:
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