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