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Unit 3

Unit 3. Vectors. Trig ReView. s in α = opposite hypotenuse cos α = adjacent hypotenuse tan α = opposite adjacent . α = tan -1 ( opp / adj ) α = cos -1 ( adj / hyp ) α =sin -1 ( opp / hyp ). hypotenuse. opposite. α ). adjacent.

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Unit 3

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  1. Unit 3 Vectors

  2. Trig ReView sinα = opposite hypotenuse cosα= adjacent hypotenuse tanα= opposite adjacent α= tan-1(opp/adj) α= cos-1(adj/hyp) α=sin-1(opp/hyp) hypotenuse opposite α) adjacent SOH CAH TOA Pythagorean Theorem: a2 + b2 = c2 *****Set calculators to degree mode******

  3. Describing Vectors • 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. Try these examples Give directions for the following in Compass, Geometric, and Bearing readings Then complete trig review wkst

  5. VECTORS • 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 distance speed displacement time velocity mass acceleration WORK FORCE ENERGY MOMENTUM

  7. Vector addition • 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

  9. Practice problems

  10. 2D • When finding displacement or avg. velocity in 2D, we need to remember the following equations: • ∆x = xf-xi ∆v = vf-vivavg = ∆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 displacement 90 m [S] • The man’s average velocity 90 m [S]/30s = 3m/s [S] 40 m 50 m

  11. Example b • 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 a2 + b2 = c2 • Average velocity (∆r)2 = (40+53km)2 + (5-3km)2 ∆r = 93.02 km vavg = ∆x/∆t = 93.02 km/2min = 46.51 km/min Dr

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

  14. 150 sin(37) = 90.3 km/h 150 km/h 150 cos(37) = 119.8 km/h 37o 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. Crooked vector addition • 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 = -63.3 Y: 0 - 25 = -25 (0 because the 1st 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 20o N] 12 ft [S 43.8o E] 12 ft @ 46.2o S of E 7 m/s @ 20o N of E

  18. 63.3 25

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

  20. In class examples • 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. 76.60 N 187.94 N 68.40 N 100 N 64.28 N 50o 20o 45 N 200 N x: y: -45 – 76.60 + 187.94 = 66.34 N 0 + 64.28 – 68.40 = -4.12 N 66.34 N FNET = 66.47 N [E 3.55o S] 4.12 N FNET

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

  23. Fy 80 N F3 Fx ???o 45o 5 N 42 N 29.70 N F3 114.7 N 29.70 N x: y: -29.70 + 0 + Fx = 0  Fx = 29.70 N -29.70 - 5 + Fy = 80  Fy = 114.7 F3 = 118. 5 N [E 75.5o N] 29.70 N

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

  25. Equilibrium problems • 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. 13.89 10o F4 = ? 80 N 78.78 15.67 N 50o 40 N 41.78 65 N F4 49.79 28.99 N x: y: -41.78 + 40 – 13.89 + Fx = 0  Fx = 15.67 N -49.79 + 0 + 78.78 + Fy = 0  Fy = -28.99 N F4 = 32.95 N [E 61.6o S]

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

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

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

  30. Vf= vi= a= Dv / Dt = 55mph[] / 30ms = 55mph[] / .030sec = 55mph[] / (8.333E-6h) = 6,600,000 mi/h2 [] 30 mph 25 mph = 814.81 m/s2 []

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

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

  33. A man is standing on a riverboat next to his wife. The boat is moving down a river without propelling itself, using only the current’s speed, which is 50 ft/min. The man starts jogging towards the front of the boat with a speed of 100 ft/min. During his jog, a smaller boat, traveling at 150 ft/min (relative to the water), passes the boat (in the same direction). A bird is sitting on the shore, watching the whole situation unravel. 150 ft/min  • 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? 100 ft/min  50 ft/min  50 ft/min  150 ft/min 

  34. So, what was the velocity of the man? It depends. What is the velocity relative to? It depends on the “Frame of Reference” of the observer.

  35. Two cars are moving towards each other on a highway. Car A moves East at 60 mph, while car B moves West at 50 mph. Find the velocity of car A with respect to …. a) car B b) a bird sitting on the side of the highway c) a boy in the backseat of car A 110 mph [E] 60 mph [E] 0 mph V BG = 50 mph V AG = 60 mph

  36. In relative velocity problems, we often use what is known as a “relative velocity equation”. When you see a term like “VBC” this means “the velocity of the Ball relative to the Car” (for example)

  37. The relative velocity equation vBS v S vB vBS vBS=vBC + vCS Here’s an easy trick to remember the equation vBS= + C C VBS = velocity of the ball relative to the side of the road VBC = velocity of the ball relative to the car VCS = velocity of the car relative to the side of the road

  38. * REALLY IMPORTANT NOTE

  39. Two cars are moving towards each other on a highway. Car A moves East at 60 mph, while car B moves West at 50 mph. Find the velocity of car A with respect to …. a) car B b) a bird sitting on the side of the highway c) a boy in the backseat of car A Looking at this problem again  110 mph [E] 60 mph [E] 0 mph 60 mph 50 mph 60 mph 0 VAB = VAG + VGB = + VAbird = VAG + VGbird = + VAboy = VAG + VGboy = + 60 mph 60 mph

  40. Example:A boy sits in his car with a tennis ball. The car is moving at a speed of 20 mph. If he can throw the ball 30 mph, find the speed with which he will could hit… • 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 30 mph 50 mph vBS=vBC + vCS 30 mph 20 mph 10 mph 20 mph vBS=vBC + vCS 30 mph

  41. River Problems

  42. River Problems Crossing a river: aiming straight across, but not accounting for the moving water  VCW = velocity of the canoe relative to the water (the velocity that the canoeist paddles) VWB = velocity of the water relative to the riverbank (the velocity of the river current) VCB = velocity of the canoeist relative to the riverbank vCB=vCW + vWB

  43. VCW = velocity of the canoe relative to the water (the velocity that the canoeist paddles) VWB = 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. VCB = velocity of the canoeist relative to the riverbank 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.

  44. Example: A 20 m wide river flows at 1.5 m/s. A boy canoes across it at 2 m/s relative to the water.a. What is the least time she requires to cross the river? b. How far downstream will she be when she lands on the opposite shore? c. What will her velocity relative to the shore be as she crosses? d = rt  20m = (2 m/s) t  t = 10 seconds. d = rt = (1.5 m/s)(10 sec) = 15 meters 1.5 2 2.5 2.5 m/s [across 36.87o downstream]

  45. River Problems Crossing a river: trying to go from point A to point B B VGW = velocity of the girl relative to the water (the velocity that the girl swims) VWB = velocity of the water relative to the riverbank (the velocity of the river current) VGB = velocity of the girl relative to the riverbank A vGB=vGW + vWB

  46. Example: A river is 20 m wide. It flows at 1.5 m/s. If a girl swims at a speed of 2 m/s, find:a) the time required for the girl to swim 15 m upstream.b) the time required for the girl to swim 20 m downstream.c) the angle (between the swimmers path and the shore that the girl should aim when crossing the river if she wants to arrive at the other side directly across from her starting point. 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 1.5 q q = cos-1 (1.5/2) = 41.4o 2 1.32 [upstream 41.4o across] q

  47. From previous example: A river is 20 m wide. It flows at 1.5 m/s. The girl swims at a speed of 2 m/s. d) How long will it take to cross the river in this case (the case where the girl adjusts for the current by pointing herself upstream) d = rt 20m = (1.32 m/s)t  t = 15.2 seconds 1.5 2 1.32 q

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

  49. Plane Problems

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