1. Describe the motion in the following illustrations Ticker tapes reveal data on which two kinematic variables?

2. Match each d-t graph to the proper v-t & a-t graph

4. Review yesterdays lesson with 3 practice problems Review yesterdays lesson with 3 practice problems SPH 3U Grade 11 U Physics 2010

5. Motion in Two Dimensions – Problem Set

6. 1 2 3

10. Now we have two perpendicular vectors: make a right triangle!

11. Now we have two perpendicular vectors: make a right triangle!

19. How did we do?

20. Time to cross river? Isolate for t What's the formula for v?

21. Time to cross river? Isolate for t ! What's the formula for v?

22. Time to cross river?

24. Time to cross depends on y -velocity only “Current” or x -velocity just pushes you downstream as you cross

25. How far downstream?

27. Resultant velocity?

34. Lets look at our 5 vector problem from yesterday – Ie. 10m N30ºE = V1 13m E15ºN = V2 21m E83ºN = V3 20m W45ºN = V4 11m N72ºE = V5

36. r y r y 50.4 km [N] r x r x 16.42 km [E] 53m E 72 N 53m E 72 º N 53m53m 72 72 º 50.416.4

37. Split up into 3 groups Create a multi-vector (2d) word problem – With solution Exchange your problem with the group to your right

38. An acrobat is launched from a cannon at an angle of 60 above the horizontal. He is caught by a safety net mounted horizontally at the height from which he was initially launched. Suppose he is launched at 26 m/s. a)How long does it take before he reaches his max height b)How long does it take in total for him to reach a point halfway back down to the ground.

39. G R A S ΔtΔt

40. G R A S ΔdyΔdy Half or mid-height would be 13m – would occur twice

41. G R A S

55. ; Δt = 5.1 s

57. Now that we know how long the object will be air born for (time of flight) it is easy to determine the range (horizontal displacement). All we need to do is take the magnitude of our vector (in this case 25m/s) and determine the horizontal or x component of it  using our trig ratio! Now that we know how much of that initial 25 m/s is moving horizontally per second, we substitute in our time of flight to see how far it will travel.

58. ; Δt = 5.1 s

62. 16.1 30.9

63. 16.1 30.9

64. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

65. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

66. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

67. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

68. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

69. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9 Since we have uniform acceleration (via gravity) we can use 1 of our 5 acceleration formula’s…. We must determine which variable we are missing and select our formula accordingly. In this case equation 4!

70. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

71. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

72. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

73. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

74. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

75. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9 Similar to displacement vectors, we can now use Pythagorean theorem to determine the resultant vector of this right angle triangle. This is the convenient thing about using the component vector process – by breaking every movement (whether it be a velocity or a displacement) into its sub-components (i.e. a vertical and a horizontal) we can determine the resultant vector rather easily. In this case, as the diagram to the right highlights… we have the final velocity in the horizontal direction and the final velocity in the vertical direction… by determining the resultant vector we should get the overall final velocity and the angle at which it is travelling

76. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

77. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

78. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

79. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

80. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

81. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9

82. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9 We can check out answer by using our trig ratio’s to see if we still get the same horizontal and vertical sub- components Vfx = 35 cos62 Vfx = 16.4 m/s Vfy = 35 sin62 Vfy = 30.9 m/s

83. ; Δt = 5.1 s ; Δd x = 82 m 16.1 30.9