1. + Two-Dimensional Motion and Vectors Chapter 3 pg. 83-109

2. + Objective(s) 1. Distinguish between a scalar and a vector. 2. Add and subtract vectors using the graphical method. 3. Multiply and divide vectors by scalars

3. + What do you think? How are measurements such as mass and volume different from measurements such as velocity and acceleration? How can you add two velocities that are in different directions?

4. + Introduction to Vectors Scalar - a quantity that has magnitude but no direction Examples: volume, mass, temperature, speed Vector - a quantity that has both magnitude and direction Examples: acceleration, velocity, displacement, force

5. + Vector Properties Vectors are generally drawn as arrows. Length represents the magnitude Arrow shows the direction Resultant - the sum of two or more vectors Make sure when adding vectors that You use the same unit Describing similar quantities

6. + Finding the Resultant Graphically Method Draw each vector in the proper direction. Establish a scale (i.e. 1 cm = 2 m) and draw the vector the appropriate length. Draw the resultant from the tip of the first vector to the tail of the last vector. Measure the resultant. The resultant for the addition of a + b is shown to the left as c.

7. + Vector Addition Vectors can be moved parallel to themselves without changing the resultant. the red arrow represents the resultant of the two vectors

8. + Vector Addition Vectors can be added in any order. The resultant (d) is the same in each case Subtraction is simply the addition of the opposite vector.

9. Sample Resultant Calculation A toy car moves with a velocity of.80 m/s across a moving walkway that travels at 1.5 m/s. Find the resultant speed of the car.

10. + Practice Time Section 3.1 Review pg. 87 Section 3.1 Diagram Skills Worksheet

11. + FOCUS QUESTION(S) Page 87 #1-3 For questions 2 and 3 just draw the vectors

12. + 3.2 Vector Operations

13. + Objective(s) 1. Identify appropriate coordinate systems for solving problems with vectors. 2. Apply the Pythagorean theorem and tangent function to calculate the magnitude and direction of a resultant vector. 3. Resolve vectors into components using the sine and cosine functions. 4. Add vectors that are not perpendicular.

14. + What do you think? What is one disadvantage of adding vectors by the graphical method? Is there an easier way to add vectors?

15. + Vector Operations Use a traditional x-y coordinate system as shown below on the right. The Pythagorean theorem and tangent function can be used to add vectors. More accurate and less time-consuming than the graphical method

16. + Pythagorean Theorem and Tangent Function

17. + We can use the inverse of the tangent function to find the angle. θ = tan -1 (opp/adj) Another way to look at our triangle d 2 = Δ x 2 + Δ y 2 d ΔyΔy ΔxΔx θ

18. + Example An archaeologist climbs the great pyramid in Giza. The pyramid height is 136 m and width is 2.30 X 10 2 m. What is the magnitude and direction of displacement of the archaeologist after she climbs from the bottom to the top?

19. + Example Given: Δ y = 136m width is 2.30 X 10 2 m for whole pyramid So, Δ x = 115m Unknown: d = ?? θ = ??

20. + Example Calculate: d 2 = Δ x 2 + Δ y 2 d = √ Δ x 2 + Δ y 2 d = √ (115) 2 +(136) 2 d = 178m θ = tan -1 (opp/adj) θ = tan -1 (136/115) θ = 49.78°

21. + Example(You Try this) While following the directions on a treasure map a pirate walks 45m north then turns and walks 7.5m east. What single straight line displacement could the pirate have taken to reach the treasure?

22. + Focus Question(s) 10/9 Cheetahs are, for short distances, the fastest land animals. In the course of a chase, cheetahs can also change direction very quickly. Suppose a cheetah runs straight north for 5.0 s, quickly turns, and runs 3.00 × 10 2 m west. If the magnitude of the cheetah’s resultant displacement is 3.35 × 10 2 m, what is the cheetah’s displacement and velocity during the first part of its run?

23. + Resolving Vectors Into Components

24. + Resolving Vectors into Components Component: the horizontal x and vertical yparts that add up to give the actual displacement For the vector shown at right, find the vector components v x (velocity in the x direction) and v y (velocity in the y direction). Assume that the angle is 35.0˚. 35°

25. + Example Given: v= 95 km/h θ = 35.0° Unknown v x =?? v y = ?? Rearrange the equations sin θ = opp/ hyp opp=(sin θ ) (hyp) cos θ = adj/ hyp adj= (cos θ )(hyp)

26. + Example v y =(sin θ )(v) v y = (sin35°)(95) v y = 54 km/h v x = (cos θ )(v) v x = (cos 35°)(95) v x = 78 km/h

27. + Example( You Try This one) How fast must a truck travel to stay beneath an airplane that is moving 105 km/h at an angle of 25° to the ground?

28. + Classwork and HW Problem 3A, 3B, 3C HW Practice 3A, 3B and 3C in the book. Pages 91,94 and 97

29. + FOCUS QUESTION(S) U.S. Highway 212 extends 55 km at 37° north of east between Newell and Mud Butte, South Dakota. It then continues for 66 km nearly due east from Mud Butte to Faith, South Dakota. If you drive along this part of U.S. Highway 212, what will be your total displacement?

30. + And another A bullet traveling 850 m ricochets from a rock. The bullet travels another 640 m, but at an angle of 36° from its previous forward motion. What is the resultant displacement of the bullet?

31. + Another Wrigley Field is one of only three original major- league baseball fields that are still in use today. Suppose you want to drive to Wrigley Field from the corner of 55th Street and Woodlawn Avenue, about 14 miles south of Wrigley Field. Although not the fastest or most direct route, the most straightforward way to reach Wrigley Field is to drive 4.1 km west on 55th Street to Halsted Street, then turn north and drive 17.3 km on Halsted until you reach Clark Street. Turning on Clark, you will reach Wrigley Field after traveling 1.2 km at an angle of 24.6° west of north. What is your resultant displacement?

32. + And another An Arctic tern flying to Antarctica encounters a storm. The tern changes direction to fly around the storm. If the tern flies 46 km at 15° south of east, 22 km at 13° east of south, and finally 14 km at 14° west of south, what is the tern’s resultant displacement?

33. + HW question(s)??

34. + 3.3 Projectile Motion

35. + What do you think? Suppose two coins fall off of a table simultaneously. One coin falls straight downward. The other coin slides off the table horizontally and lands several meters from the base of the table. Which coin will strike the floor first? Explain your reasoning. Would your answer change if the second coin was moving so fast that it landed 50 m from the base of the table? Why or why not?

36. + Projectile Motion Projectiles: objects that are launched into the air tennis balls, arrows, baseballs, javelin Gravity affects the motion Projectile motion: The curved path that an object follows when thrown, launched or otherwise projected near the surface of the earth

37. + Projectile Motion Path is parabolic if air resistance is ignored Path is shortened under the effects of air resistance

38. Components of Projectile Motion As the runner launches herself (v i ), she is moving in the x and y directions.

39. + Projectile Motion Projectile motion is free fall with an initial horizontal speed. Vertical and horizontal motion are independent of each other. Vertically the acceleration is constant (-9.81 m/s 2 ) We use the 4 acceleration equations Horizontally the velocity is constant We use the constant velocity equations

40. + Projectile Motion Components are used to solve for vertical and horizontal quantities. Time is the same for both vertical and horizontal motion. Velocity at the peak is purely horizontal (v y = 0).

41. Vertical and Horizontal Motion Horizontal MotionVertical Motion Velocity = V x Displacement = Δx Velocity = V y Displacement = Δy Because gravity does not act in the horizontal direction, V x is always constant! Gravity acts vertically, therefore a = -9.81 m/s 2

42. Equations for projectiles launched horizontally Horizontal MotionVertical Motion Δx= V x t V x is constant! a=0 V y,i =0 (initial velocity in y direction is 0)

43. Revised Kinematic Eqns for projectiles launched horizontally Horizontal Motion Vertical Motion

44. + Finding the total velocity Use the pythagorean theorem to find the resultant velocity using the components (V x and V y ) Use SOH CAH TOA to find the direction Vx Vy V

45. + Example p. 102 #2 A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. What was the cat’s speed when it slid off the table?

46. + What do we know and what are we looking for? 2.2m Δ x= 2.2 m Δ y= -1.0m (bc the cat falls down) 1.0 m Vx= ????? What are we looking for??

47. How do we find Vx? Equation for horizontal motion: We have x…so we need t. How do we find how long it takes for the cat to hit the ground? Use the vertical motion kinematic equations.

48. Vertical Motion Δ y= -1.0m a=-9.81 m/s 2 What equation should we use? Rearrange the equation, to solve for t then plug in values.

49. Horizontal equation Rearrange and solve for Vx: Cat’s Speed is 4.9 m/s

50. + Cliff example A boulder rolls off of a cliff and lands 6.39 seconds later 68 m from the base of the cliff. What is the height of the cliff? What is the initial velocity of the boulder?

51. How high is the cliff? Δ y= ?a=-9.81 m/s 2 t = 6.39 sV x =? V y,i = 0 Δ x= 68 m The cliff is 200 m high

52. What is the initial velocity of the boulder? The boulder rolls off the cliff horizontally Therefore, we are looking for Vx

53. Important Concepts for Projectiles Launched Horizontally Horizontal ComponentsVertical Components Horizontal Velocity is constant throughout the flight Horizontal acceleration is 0 Initial vertical velocity is 0 but increases throughout the flight Vertical acceleration is constant: -9.81 m/s 2

54. + Example The Royal Gorge Bridge in Colorado rises 321 m above the Arkansas river. Suppose you kick a rock horizontally off the bridge at 5 m/s. How long would it take to hit the ground and what would it’s final velocity be?

55. + Example Given: d = 321ma = 9.81m/s 2 v i = 5m/st = ??v f = ?? REMEMBER we need to figure out : Up and down aka free fall (use our 4 acceleration equations) Horizontal (use our constant velocity equation)

56. + FOCUS QUESTION(S) (Horizontal Launch) People in movies often jump from buildings into pools. If a person jumps horizontally by running straight off a rooftop from a height of 40.0 m to a pool that is 10.0 m from the building, with what initial speed must the person jump? Answer:3.50m/s

57. + Projectiles Launched at an Angle We will make a triangle and use our sin, cos, tan equations to find our answers Vy = V sin θ Vx = V cos θ tan = θ (y/x)

58. Projectiles Launched at An Angle Projectiles Launched Horizontally Projectiles Launched at an Angle V x is constant Initial V y is 0 V x is constant Initial V y is not 0 ViVi V x,i V y,i θ V i = V x

59. Components of Initial Velocity for Projectiles Launched at an angle Use soh cah toa to find the V x,i and V y,i ViVi V x,i V y,i θ

60. Revise the kinematic equations again Horizontal MotionVertical Motion

61. + Example p. 104 #3 A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high was the tallest spot in the ball’s path?

62. What do we know? Δx= 42.0 m θ= 25° V i = 23.0 m/s V y at top = 0 Δt=? Δy=? 42.0 m 25°

63. What can we use to solve the problem? Find t using the horizontal eqn: Δ x=v x Δ t = v i cos( θ )t How to find Δ y? V y,f = 0 at top of the ball’s path What equation should we use?

64. + Focus Question(s) The narrowest strait on earth is Seil Sound in Scotland, which lies between the mainland and the island of Seil. The strait is only about 6.0 m wide. Suppose an athlete wanting to jump “over the sea” leaps at an angle of 35° with respect to the horizontal. What is the minimum initial speed that would allow the athlete to clear the gap? Neglect air resistance.

65. + Cliff example(FOCUS QUESTIONS) A girl throws a tennis ball at an angle of 60°North of East from a height of 2.0 m. The ball’s range is 90 m and it is in flight for 6 seconds. What is the initial horizontal velocity of the ball? What is the initial vertical velocity of the ball? What is the total initial velocity of the ball? How high above the initial position does the ball get? What is the vertical velocity of the ball 2 seconds after it is thrown?

66. What is the initial horizontal velocity of the ball? Δ x= 90 m Θ =60° Total time= 6 s Horizontal velocity is constant: Vx

67. + What is the initial vertical velocity of the ball? Vx,i Vy,i θ Vi

68. + What is the total initial velocity of the ball? Vx,i Vy,i θ Vi

69. How high above the ground does the ball get? At the top of the parabola, Vy is 0…so use the revised kinematic equations Add 2m to get the height above the ground: 36.65 m

70. What is the vertical velocity of the ball 2 seconds after it is thrown? V y,i =+26 m/s a= -9.81 m/s 2 t = 2 seconds

71. + Important Concepts for Projectiles Launched at an Angle At the top of the parabola, neither the object’s velocity nor it’s acceleration is 0!!!!! Only V y is 0 V x is constant throughout the flight Horizontal acceleration is always 0 Vertical acceleration is always -9.81 m/s 2

72. + Classroom Practice Problem (Projectile Launched at an Angle) A golfer practices driving balls off a cliff and into the water below. The edge of the cliff is 15 m above the water. If the golf ball is launched at 51 m/s at an angle of 15°, how far does the ball travel horizontally before hitting the water? Answer: 1.7 x 10 2 m (170 m)

73. + FOCUS QUESTION(S) A snowball is launch with the initial velocity 25 m/s at an angle of 37 ° the horizontal, how long will it take the snowball to reach its maximum height above the ground, What is the maximum height, h, above the ground that a snowball reaches after it has been launched, and What is the horizontal distance, x, the snowball has traveled when it reaches its maximum height?

74. + GET OUT YOUR HOMEWORK (BOTH VOCAB and 3F)

75. + 3.4 Relative Motion

76. + Relative Velocity

77. + 3.4 Relative Velocity Who is “at rest” in this picture? Is it the ground-based observer? The seated passengers in the train? As the earth turns and the universe expands, one thing is certain: the idea of velocity is relative.

78. + 3.4 Relative Velocity Suppose that the ground-based observer and the seated passengers in the train have an attached coordinate axis. Each is “at rest” in their coordinate system. This means the origin of their respective coordinate systems is always the same distance away. A coordinate system in which an experimenter can make position and time measurements (with help if necessary!) is called a reference frame. The ground-based observer, the seated train passengers and the walking passenger each have a different reference frame. All reference frames in this chapter have constant velocity (inertial reference frames).

79. + 3.4 Relative Velocity Consider the reference frame of the train (and its seated passengers). Let v PT = velocity of the (moving) Passenger relative to the Train = +2.0 m/s Consider the reference frame of the ground-based observer. Let V TG = velocity of the Train relative to the Ground = +9.0 m/s Then what is the velocity of the Passenger relative to the Ground ( V PG )?

80. + Relative Velocity A person, standing on the ground, measures the velocity of a car to be V C G where: V C G = velocity of the Car relative to the Ground = +25 m/s. The driver of the car looks out the window and sees the ground go past her with a velocity of V GC, where: V G C = velocity of the Ground relative to the Car. How is V G C related to V C G ? A.V G C = V C G = +25 m/s B.V G C = -V C G = -25 m/s C.V GC = 0 since the ground isn’t moving

81. + A car and a bus A slower moving car is traveling behind a faster moving bus. In the reference frame of a ground based observer, the velocities of the two vehicles are: v CG = velocity of the Car in the reference frame of the Ground = +15 m/s v BG = velocity of the Bus in the reference frame of the Ground = +20 m/s. What is the velocity of the bus in the reference frame of the car? if the In other words, what is the velocity of the bus as measured by the driver of the car? A.+35 m/s B.+5 m/s C.-35 m/s D.-5 m/s

82. + A car and a bus A slower moving car is traveling behind a faster moving bus. In the reference frame of a ground based observer, the velocities of the two vehicles are: v CG = velocity of the Car in the reference frame of the Ground = +15 m/s. v BG = velocity of the Bus in the reference frame of the Ground = +20 m/s. What is the velocity of the bus in the reference frame of the car? if the In other words, what is the velocity of the bus as measured by the driver of the car? V BC = V BG + V GC where V GC = - V CG = -15 m/s V BC = +20 m/s + - 15 m/s = +5 m/s

83. + Up the down escalator The escalator that leads down into a subway station has a length of 30.0 m and a speed of 1.8 m/s relative to the ground. A student is coming out of the station by running in the wrong direction on this escalator. The local record time for this trick is 11 s. Relative to the escalator, what speed (magnitude only) must the student exceed in order to beat the record? If we follow the rule, our equation should be: V SE = V SG + V GE

84. + Up the down escalator The escalator that leads down into a subway station has a length of 30.0 m and a speed of 1.8 m/s relative to the ground. A student is coming out of the station by running in the wrong direction on this escalator. The local record time for this trick is 11 s. V SE = V SG + V GE Knowns V EG = +1.8 m/s: magnitude is given, sign is arbitrary V GE = - V EG = -1.8 m/s V SG = - (30 m/11s) = - 2.7 m/s: relative to the ground- based observer, the student traveled 30 m in 11 s. Since he walked against the elevator, the sign is opposite to that of the elevator relative to the ground. Find: |V SE |, speed of the student relative to the elevator.

85. + Up the down escalator The escalator that leads down into a subway station has a length of 30.0 m and a speed of 1.8 m/s relative to the ground. A student is coming out of the station by running in the wrong direction on this escalator. The local record time for this trick is 11 s. Relative to the escalator, what speed (magnitude only) must the student exceed in order to beat the record? V SE = V SG + V GE V SE = (-2.7 + - 1.8)m/s V SE = -4.5 m/s speed is 4.5 m/s

86. + 3.4 Relative Velocity Example 11 Crossing a River The engine of a boat drives it across a river that is 1800m wide. The velocity of the boat relative to the water is 4.0m/s directed perpendicular to the current. The velocity of the water relative to the shore is 2.0m/s. (a) What is the velocity of the boat relative to the shore? (b) How long does it take for the boat to cross the river?

87. + 3.4 Relative Velocity

88. +

89. + Practice time Practice 3 F and section Review

90. + FOCUS QUESTION(S) 1. A plane can travel with a speed of 80 mi/hr. Determine the resultant velocity of the plane (magnitude only) if it encounters a a. 10 mi/hr headwind. b. 10 mi/hr tailwind. c. 10 mi/hr crosswind. d. 60 mi/hr crosswind.

91. + MORE FOCUS QUESTION(S) A motorboat traveling 4.0 m/s, East encounters a current traveling 7.0 m/s, North. a. What is the resultant velocity of the motorboat? b. If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore? c. What distance downstream does the boat reach the opposite shore?

92. + Practice Time COMPLETE 3.4 Diagram skills and 3.4 section Review on page 109

93. + FOCUS QUESTION(S) An ant on a picnic table travels 31 cm eastward, then 25 cm northward, and finally 15 cm westward. What is the ant’s directional displacement relative to its original position?

94. + WHO WANTS TO BE HW COUPONS MILLIONAIRE