PROJECT 1
DIVING DECEPTION PASS
• Deception Pass is a narrow channel
between two large islands near Seattle,
Washington. There is a bridge across the
channel that connects the two islands.
When the tide comes in, the entire Pacific
Ocean pushes water throw that channel to
get to the inner part of the Puget sound.
EQUATIONS
v0
x' 
 1
S ( x)
1
S ( x)  1 
( x  100) 2  2
200000
v0 sin(  t / 23400)
x' 
 3
S ( x)
1
1
S ( x)  3 
x  600 
x  4
300
1000
v0 sin t / 23400
x' 
 5
1
1
3
x  600 
x
300
1000
PROBLEM 1
• Suppose that the velocity V0 of the current
under the bridge is 7 mph. Convert V0 to
feet per second, and then find the Velocity
of the current 600 feet to the west of the
bridge. Likewise find the velocity of the
current 200 feet east of the bridge, and
800 feet east of the bridge.
SOLUTION 1
v0  7mph  10.27 ft / sec .
v0
x' 
1
Case I
x' 
v0
1
1
( x  100) 2
200000

1
( x  100) 2
200000
10.27
1
1
(600  100) 2
200000
 2.97681 ft/sec.
Case II
v0
x' 
1
Case III
x' 
1
1
( x  100) 2
200000
v0
1
( x  100) 2
200000
10.27

1
1
(200  100) 2
200000
10.26

1
1
(200  100) 2
200000
 9.78095 ft/sec.
 2.97681 ft/sec.
PROBLEM 2
• Under the assumptions of Problem 1,
where is the current faster?
SOLUTION 2
x' 
x' ' 
1
1
1
( x  100) 2
200000
10
0.0001027( x  100)
1

2 
1

(
x

100
)
)
 200000

2
8
6
x' '  0

0.0001027( x  100)
1

2 
1

(
x

100
)
)
 200000


 0.0001027( x  100)  0
 x  100 ft
2
4
0
2
-1500
-1000
-500
500
1000
1500
PROBLEM 3 (CD)
• You will learn how to solve differential
equations such as (1) in Chapter 2. For
the moment, use the Deception Pass Tool
on the DE Tools CD to see how your
motions changes depending on where you
start. In words, describe your motion if you
enter the channel 1000 feet west of the
bridge. Describe your motion if you enter
1000 feet east of the bridge.
SOLUTON 3
• We can see that there is acceleration in
the motion until it reaches a maximum
speed of 10.27 ft/sec, then the motion
starts to decelerate.
10
0.015
8
0.01
6
0.005
-1500
4
-1000
-500
500
-0.005
2
-0.01
-1500
-1000
-500
500
1000
1500
-0.015
1000
1500
PRPBLEM 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.
SOLUTION 4
x' 
v0 sin(  t / 23400)
1
1
( x  100) 2
200000
v0 sin(  t / 23400)
x' 
1
1
( x  100) 2
200000
13.5  sin(  (3600) / 23400)
x' 
1
1
(1000  100) 2
200000
x'  0.889895ft /sec
PROBLEM 5
• You want to compute your likely position
relative to the bridge from the time you
entered the water (exactly 1 hour after
slack water, at a point 1000 feet west of
the bridge). What initial-value problem
would you solve to do so?
SOLUTION 5
v0 sin(  t / 23400)
1
1
( x  100) 2
200000
v0 sin(  t / 23400)
dx

1
dt
1
( x  100) 2
200000
1
1
( x  100) 2 dx  v0 sin(  t / 23400)dt
200000
1
2
1

(
x

100
)
dx   v0 sin(  t / 23400)dt
 200000
  t 
23400  v0  cos 
2
3
21x
x
x
23400 




 C  IVP
20 2000 600000

x' 
at t  3600 sec .  x  1000
PROBLEM 6
• Show that the model (5) 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 DE in view of the sign of the
channel-size function S(x).
SOLUTION 6
x' 
v0 sin t / 23400
1
1
3
x  600 
x
300
1000
Case I (x= - 300)
x' 
v0 sin t / 23400 
12.5 sin  (3600) / 23400 

 2.72772ft/ sec.
1
1
1
1
3
x  600 
x 3
 300  600 
(300)
300
1000
300
1000
Case II (x= 300)
x' 
v0 sin t / 23400
12.5 sin  (3600) / 23400

 -20.9125ft/sec.
1
1
1
1
3
x  600 
x 3
300  600 
(300)
300
1000
300
1000
That’s means that somewhere in the in the
interval of [-300,300] the direction of flow will
be reversed by it self which is impossible.
That means that the channel-size function is
not valid
50
-1000
-500
500
-50
-100
1000
Thank You
Muchas Gracias