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Crossing the River Discussion

Crossing the River Discussion. Created for CVCA Physics by Dick Heckathorn 15 October 2K+4. Q-1 Problem 1 Heading downstream (w/current) v = ?. Q-2 Heading upstream (against current) V=?. Q-3 What is shore doing?. Q-3a Tuber observation of boat.

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Crossing the River Discussion

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  1. Crossing the RiverDiscussion Created for CVCA Physics by Dick Heckathorn 15 October 2K+4

  2. Q-1 Problem 1 Heading downstream (w/current) v = ? Q-2 Heading upstream (against current) V=? Q-3 What is shore doing? Q-3a Tuber observation of boat Q-4 Heading across river (no current) v = ? Q-5 Heading across river (w/current) v = ? Q- Study the v and d vector drawings Q-6 Going directly across the river with current Quiz 2 Quiz 4 Quiz 1 Quiz 3

  3. Problem We wish to cross a river. This unique river has a constant current of 6 m/s from bank to bank. Speed of the Current 6 m/s

  4. We will be crossing in a boat that can travel with a speed of 8 m/s in still water. (This means that the boat will have this speed in the direction that the boat is pointing.) Speed of the boat 8 m/s

  5. Question 1Heading Downstream If the boat is heading downstream, what will be the resultant speed of the boat as seen from shore? VC = 6 m/s VB = 8 m/s VR = 14 m/s

  6. Question 2Heading Upstream If the boat is heading upstream, what will be the resultant speed of the boat as seen from shore? VR = 2 m/s VC = 6 m/s VB = 8 m/s

  7. Question 3Changing to Boat Reference Frame How fast will a person in the boat see the shore pass by? At the same speed that the person on shore sees the boat pass by - but - in the opposite direction.

  8. Question 3aChanging To Tuber Reference Frame How fast will a tuber drifting with the water see the water pass by? The tuber will be floating with the water. Thus they will say the speed of the water relative to them is zero.

  9. Question 3aChanging Reference Frame How fast will a tuber see the boat go by? VB+C = 14 m/s - VB = 8 m/s VC = 6 m/s

  10. Question 3aChanging Reference Frame What is the new velocity of the boat? Has the velocity of the boat changed? I trust you said no. Thus the tuber always sees the boat traveling at a speed of 8 m/s and... with a velocity of 8 m/s in the direction it is heading. (downstream)

  11. Problem 4Heading Across - No Current The river has no current. We will now head (in the boat) across the river which is 120 m wide.

  12. Problem 4Heading Across - No Current he crossing displacement is: he velocity of the boat is: dA = 120m VB = 8 m/s Note…Both must have different scales!

  13. Problem 4Heading Across - No Current How long will it take to cross the river? We know : VB = 8 m/s dA = 120m

  14. Problem 5Heading Across - With Current Now lets turn the current back on and head directly across the river. What will be the resultant velocity of the boat?

  15. Problem 5 Heading Across - With Current Then find the magnitude of the resultant: Finally, find the angle . First draw the velocity vector of the boat. Next add the current vector. VC = 6 m/s Next, draw the resultant vector Thus the resultant velocity of the boat is VR = 10 m/s VB = 8 m/s 36.9o 10 m/s  across 36.9o downstream

  16. Problem 5 Heading Across - With Current How long will it take to cross the river? We know that: VC = 6 m/s d = 120 m VB = 8 m/s Which v ? VR = 10 m/s 10 m/s

  17. Problem 5 Heading Across - With Current but 12 sec is incorrect. Nice try… Why? Where do we go from here? We have learned that a vector times a scalar results in a vector with the same direction as the original vector. Thus when we divide a vector by a vector, they both must be in the same direction.

  18. Problem 5 Heading Across - With Current We know that: VC = 6 m/s d = 120 m VB = 8 m/s Which v ? VR = 10 m/s 8 m/s

  19. Problem 5 Heading Across - With Current Are you surprised that the time (15 sec) to cross the river heading at right angle to the current is the same time that it took to cross the river with no current? (15 sec)?

  20. Problem 5 Heading Across - With Current If so, remember that the current is at right angles to the direction you are heading. Thus the current has no affect on the crossing speed or the crossing time.

  21. Problem 5Heading Across - With Current How far downstream will we be when we reach the other shore? What time and which velocity will we use? The one that is in the same direction as the displacement which we want to find. This equation relates distance to velocity and time.

  22. Problem 5 Heading Across - With Current And the answer is: VC = 6 m/s t = 15 s

  23. Study The Following Vc = 6 m/s dD = 90m vB = 8 m/s dA = 120m vR = 10 m/s dR = 150m Thus different scales must be used for the velocity and the displacement. Each vector on the right is equal to the vector on the left multiplied by 15 sec. They are completely different vectors. How does the triangle on the right compare to the one on the left? They are similar triangles: a vector times a scalar = a vector in the same direction.

  24. Problem 6Going Directly Across with Current At what direction must one head so that we will reach the shore directly across the river from where we started?

  25. Problem 6 Going Directly Across with Current We know the following: VC = 6 m/s Speed of boat = 8 m/s Direction we wish to go.

  26. Problem 6 Going Directly Across with Current We proceed as follows: Then add the speed of the boat with its head at the tail of the current and its tail on the resultant direction. Finally, add the two vectors. First, draw the current velocity vector. Next add a dotted line indicating the direction of the resultant velocity. VC = 6 m/s VB = 8 m/s

  27. Problem 6Going Directly Across with Current What trig function can we use to find angle ? We have the side opposite and the hypotenuse. We can use the sine function to find the angle . What do you think angle  is? Next we will need to find the direction the boat will head? VC = 6 m/s Did you guess 36.9o? Sorry ? ? ? VB = 8 m/s That is not the correct answer? 

  28. Problem 6Going Directly Across with Current We will use pythagorean theorem: Finally we need to find the resultant speed of the boat. The resultant speed of the boat is VC = 6 m/s VB = 8 m/s 

  29. Problem 6Going Directly Across with Current The resultant velocity of the boat is Or we could use the vadd computer program. The velocity of the current is Giving the result of: VC = 6 m/s VB = 8 m/s  = 5.83 m/s 90o

  30. Problem 6Going Directly Across with Current The resultant velocity of the boat is: VC = 6 m/s VB = 8 m/s  across, 48.6o upstream

  31. Quiz Question 1 A boat heads directly across a river 41 m wide at 3.8 m/s. The current is 2.2 m/s. a. What is resultant velocity of the boat? b. How long to cross the river? c. How far downstream is boat when it reaches the other side?

  32. Quiz Question 1 Answers a. 4.39 m/s across 30.1o downstream b. 10.8 sec c. 23.7 m

  33. Quiz Question 1 Answers (not to scale) VC = 2.2m/s dD = VB = 3.8 m/s VR = dR = dA = 41m

  34. Quiz Question 2 A boat heads directly across a river 187 m wide at 10.3 m/s. The current is 6.7 m/s. a. What is resultant velocity of the boat? b. How long to cross the river? c. How far downstream is boat when it reaches the other side?

  35. Quiz Question 2 Answers a. 12.3 m/s across 33.0o downstream b. 18.2 sec c. 121.9 m

  36. Quiz Question 2 Answers (not to scale) VC = 6.7m/s dD = VB = 10.3 m/s VR = dA = 187m dR =

  37. Quiz Question 3 A boat heads directly across a river 110 m wide at 8.5 m/s. The current is 3.8 m/s. a. What is resultant velocity of the boat? b. How long to cross the river? c. How far downstream is boat when it reaches the other side?

  38. Quiz Question 3 A boat heads directly across a river 110 m wide at 8.5 m/s. The current is 3.8 m/s. d. At what angle will the boat head so that it will go directly across? e. What will be the crossing speed when heading directly across? f. How long to cross?

  39. Quiz Question 3 Answers a. 9.31 m/s across 24.1o downstream b. 12.9 sec c. 49.2 m d. across 26.6o upstream e. 7.60 m/s f. 14.5 sec

  40. Quiz Question 3 Answers (not to scale) VC = 3.8/s dD = VB = 8.5 m/s VR = dA = 110m dR =

  41. Quiz Question 4 A boat heads directly across a river 245 m wide with a speed of 14.2 m/s. The current is 8.3 m/s. a. What is resultant velocity of the boat? b. How long will it take the boat to cross the river? c. How far downstream is boat when it reaches the other side?

  42. d. At what angle will the boat head so that it will go directly across? e. What will be the crossing speed when heading directly across? • How long to cross? g. How long will it take the boat to cross the river if there is no current?

  43. Quiz Question 4 Answers a. 16.45 m/s across 30.3o downstream b. 17.3 sec c. 143.2 m d. across 35.8o upstream e. 11.5 m/s f. 21.3 sec g. 17.3 sec

  44. That’s all folks!

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