1. Vectors

2. α)α) hypotenuse opposite adjacent sin α = opposite hypotenuse cosα= adjacent hypotenuse tanα= opposite adjacent α= tan -1 (opp/adj) α= cos -1 (adj/hyp) α=sin -1 (opp/hyp) Pythagorean Theorem: a 2 + b 2 = c 2 *****Set calculators to degree mode****** SOH CAH TOA

3.  We use 3 different methods:  Compass (N,E,S,W) Describe the angle using two compass directions. EX: [W 25°N] Answer can also be recorded as [N 65°W] but we try to keep the degree value less than 45°  Geometric East equals 0° and you read values counterclockwise  Bearing North equals 0° and you read values clockwise

4. Give directions for the following in Compass, Geometric, and Bearing readings Then complete trig review wkst

5.  Vector-a quantity with both a magnitude (a size or number) and a direction (which way it is pointing)  Example: Velocity v= 40 m/s North 30˚ East  Scalar- a quantity with a magnitude only.  Example: Speed s= 40 m/s

6. proportional position displacement velocity acceleration FORCE MOMENTUM distance speed time mass WORK ENERGY

7.  Vectors can be added or subtracted. When they are, the result is another vector called a resultant vector.  Vectors are always added tip to tail.  To subtract vectors, add the opposite.

8. Tip-to-Tail

10.  When finding displacement or avg. velocity in 2D, we need to remember the following equations:  ∆x = x f -x i ∆v = v f -v i v avg = ∆x/∆t  Example: A certain man’s initial position is 40m [N], as measured from his house. After 30 seconds of walking, his new position is 50 m [S]. Find The distance travelled by the man during the trip 90m The man’s displacement90 m [S] The man’s average velocity 90 m [S]/30s = 3m/s [S] 40 m 50 m

11.  A radar station is tracking a jet. Its location at a certain moment on time is 40 km [W] at an altitude of 5000m, relative to the station. 2 minutes later, its location is 53 km [E], this time at an altitude of 3000m. Find the jet’s  Displacement a 2 + b 2 = c 2  Average velocity(∆r) 2 = (40+53km) 2 + (5-3km) 2 ∆r = 93.02 km v avg = ∆x/∆t = 93.02 km/2min = 46.51 km/min rr

12.  Every vector can be written as a combination of only x- direction and y-direction vectors. These vectors are called the COMPONENTS of the original vector. When we break a vector into its components, we call this RESOLVING the vector into components.

13. 10 cos(30) = 8.66 m/s 10 sin(30) = 5 m/s The bird’s shadow is moving at a speed of 8.66 m/s relative to the ground. 10 m/s 30 o

14. 150 km/h 37 o 150 cos(37) = 119.8 km/h 150 sin(37) = 90.3 km/h d = rt = (90.3 km/h)(3/3600 h) =.075 km = 75 m = y d = rt = (119.8 km/h)(3/3600 h) =.100 km = 100 m = x

15.  When vectors are “crooked”, they become slightly more complicated to add together.  Driving 20 km [W] then 50 km [W30˚S]. The crooked vector must be broken down into it’s component pieces using Pythagorean Theorem. The 50 km vector is now represented in pure x and y values. The 20 km vector is still fine. Now, add up the x values. X: -20 - 43.3 = -63.3Y: 0 - 25 = - 25 (0 because the 1 st vector had no Y value)

16. Now, recombine the x and y: 63.3 km 25 km Note that the -63.3 means the vector should point left and the -25 means the vector should point down. Next the resultant vector can be drawn in. 68.1 km Θ = Tan -1 (25/63.3) = 21.55˚ ∆d = 68.1 km [W 21.55˚S] Or 68.1 km [S 68.45 ˚W] To give the proper direction you must follow the component of the vector whose direction you are giving. The angle included in the directions is always the angle included between the “starting” component and the resultant. EX: W 21.55˚S NOT S 21.55˚W Also notice there are 2 different possible answers depending on whether we started with the south vector or the west vector.

17. 7 m/s [E 20 o N] 7 m/s @ 20 o N of E 12 ft [S 43.8 o E] 12 ft @ 46.2 o S of E

18. 63.3 25

19. 63.3 25 68.1 21.55 o 68.45 o 68.1 km [W 21.55 o S] 68.1 km [S 68.45 o W]

20.  When adding crooked vectors, use the following procedure:  Draw all the vectors appropriately  Resolve each vector into its components  Add the x-components together (making sure to keep track of the signs) yielding SUPER-X  Add the y-components together (making sure to keep track of the signs) yielding SUPER-Y  Use SUPER-X and SUPER-Y to construct a SUPER- TRIANGLE  Find the resultant (magnitude and direction) of the SUPER-TRIANGLE

21. 45 N 100 N 200 N 50 o 20 o 76.60 N 64.28 N 187.94 N 68.40 N x: y: -45 – 76.60 + 187.94 = 66.34 N 0 + 64.28 – 68.40 = -4.12 N 66.34 N 4.12 N F NET F NET = 66.47 N [E 3.55 o S]

22. 200 lb 300 lb 400 lb 50 o 20 o 281.91 lb 102.61 lb 257.12 lb 306.42 lb x: y: -281.91 + 0 + 257.12 = -24.79 lb -120.61 + 200 + 306.42 = 385.81 lb 385.81 lb 24.79 lb F NET F NET = 386.6 lb [W 86.3 o N]

23. 29.70 N F x FyFy x: y: -29.70 + 0 + F x = 0  F x = 29.70 N -29.70 - 5 + F y = 80  F y = 114.7 114.7 N 29.70 N F3F3 F 3 = 118. 5 N [E 75.5 o N] 42 N F3F3 ??? o 45 o 5 N 80 N

24. “Day #3 Vectors HW Problems” (from the packet) #’s 18-23

25.  Equilibrium problems are problems in which an object has balanced forces acting on it.  In these problems the values of the x- components of all the vectors involved sum to zero.  This is also true for the sum of the values of the y-components

26. x: y: -41.78 + 40 – 13.89 + F x = 0  F x = 15.67 N -49.79 + 0 + 78.78 + F y = 0  F y = -28.99 N 28.99 N 15.67 N F4F4 F 4 = 32.95 N [E 61.6 o S] 80 N 10 o 40 N 41.78 49.79 65 N 50 o 78.78 13.89 F 4 = ?

27. 375 lb 649.5 lb 750 lb F pole W x: y: F + 375 = 0  F = -375 lb or F = 375 lb  649.5 – W = 0  W = 649.5 lb Notice that for equilibrium problems (where nothing is moving), the components always sum up to zero. b)

28. 150 N W 70 o 140.95 N 51.30 N x: y: 140.95 – 140.95 = 0  DUH! 51.30 + 51.30 – W = 0  W = 102.6 N 140.95 N 51.30 N

29. “Day #4 Vectors HW Problems” (from the packet) #’s 24-29

30. v i = Vf =Vf = 30 mph 25 mph a  =  v /  t = 55mph[  ] / 30ms = 55mph[  ] /.030sec = 55mph[  ] / (8.333E-6h) = 6,600,000 mi/h 2 [  ] = 814.81 m/s 2 [  ]

31. vi =vi =10 m/s 20 o 9.396 m/s 3.420 m/s 3.254 m/s v f =8 m/s 24 o 7.308 m/s 2.088 m/s 6.674 m/s v f =8 m/s 24 o 3.254 m/s 7.308 m/s vi =vi =10 m/s 20 o 9.396 m/s 3.420 m/s 7 m/s @ 72.6 o off the wall in the reverse direction

33. “Day #5 Vectors HW Problems” (from the packet) #’s 30-35 QUIZ TOMORROW!!!

34. What is the man’s velocity (relative to the bird) when he is running? What is the man’s velocity (relative to his wife) when he is running? What is the man’s velocity (relative to the speedboat) when he is running? What is the wife’s velocity relative to the bird? What is the wife’s velocity relative to the speedboat? 150 ft/min  100 ft/min  50 ft/min  50 ft/min  150 ft/min 

35. It depends on the “Frame of Reference” of the observer.

36. 110 mph [E] 60 mph [E] 0 mph V AG = 60 mph V BG = 50 mph

37. When you see a term like “V BC ” this means “the velocity of the Ball relative to the Car” (for example)

38. v BS = v BC + v CS v BS v S vBvB v BS C C v BS = + V BS = velocity of the ball relative to the side of the road V BC = velocity of the ball relative to the car V CS = velocity of the car relative to the side of the road Here’s an easy trick to remember the equation

39. * REALLY IMPORTANT NOTE

40. V AB = V AG + V GB = + 60 mph 50 mph 110 mph [E] 60 mph [E] V Abird = V AG + V Gbird = + 60 mph 0 0 mph V Aboy = V AG + V Gboy = + 60 mph 60 mph Looking at this problem again

41. a)his brother in the front seat of the car. b) A sign on the side of the road that the car is about to pass. c) A sign on the side of the road that the car has already passed v BS = v BC + v CS 50 mph 30 mph20 mph 10 mph 30 mph 20 mph v BS = v BC + v CS 30 mph

43. Crossing a river: aiming straight across, but not accounting for the moving water  V CB = velocity of the canoeist relative to the riverbank V CW = velocity of the canoe relative to the water (the velocity that the canoeist paddles) V WB = velocity of the water relative to the riverbank (the velocity of the river current) v CB = v CW + v WB

44. V CB = velocity of the canoeist relative to the riverbank V CW = velocity of the canoe relative to the water (the velocity that the canoeist paddles) V WB = velocity of the water relative to the riverbank (the velocity of the river current) When answering any questions about going “ACROSS” the river, use the ACROSS THE RIVER Component. When answering any questions about going “DOWNSTREAM, use the DOWNSTREAM Component. The “slanted” vector is how the canoe will move from the perspective of anyone standing on the shore.

45. 2 1.5 2.5 d = rt  20m = (2 m/s) t  t = 10 seconds. d = rt = (1.5 m/s)(10 sec) = 15 meters 2.5 m/s [across 36.87 o downstream]

46. Crossing a river: trying to go from point A to point B V GW = velocity of the girl relative to the water (the velocity that the girl swims) V WB = velocity of the water relative to the riverbank (the velocity of the river current) B A V GB = velocity of the girl relative to the riverbank v GB = v GW + v WB

47. d = rt  15m = (2 m/s – 1.5 m/s)t  t = 30 seconds d = rt  15m = (2 m/s + 1.5 m/s)t  t = 4.29 seconds 2 1.5 1.32    = cos -1 (1.5/2) = 41.4 o [upstream 41.4 o across]

48. d = rt  20m = (1.32 m/s)t  t = 15.2 seconds 2 1.5 1.32 

49. “Day #7 Vectors HW Problems” (from the packet) #’s 36-44

51. P: Plane W: Wind G: Ground v PG = v PW + v WG “airspeed” means V PW You can use d = rt in these problems only because a = 0

52. v PG = v PW + v WG 50 45 o 400 = + 282.84 = 436.8 49.6 o V PG = 436.8 mph [E 49.6 o N]

53. x: y: V y = 226.51 – 200 = 26.51 mph Putting together the super triangle yields V WG = 47.9 mph [W 33.6 o N] 200 mph 10 o 226.51 mph 230 mph 39.94 mph ??? v WG = v PG + -v PW rearranging v PG = v PW + v WG V x = -39.94 + 0 = -39.94 mph From tonight’s HW

54. “Day #8 Vectors HW Problems” (from the packet) #’s 45-50