1. Chapter 2 Motion in One Dimension

2. Motion is relative.

3. An object can be moving with respect to one object and at the same time be at rest or moving at a different speed with respect to another.

4. Frame of Reference Is the point with which a motion is described.

5. How fast are you moving at this moment?

6. Depends upon how you look at it? If you look at it from the point of view with the room, most of you are not moving.

7. If you look at it from the point of view from outer space, then you are moving as fast as the earth is rotating.

8. Or from out side the solar system, the earth is moving around the sun at a speed of approximately 100,000 km/hr.

9. Relative Motion Animation

10. Vector A physical quantity that has both magnitude and direction.

11. Ex:10 km, North 15 m/s, SW

12. Scalar A physical quantity that has magnitude, but no direction.

13. Ex:55 km/hr 19 m

14. Distance How far an object has moved. No direction, therefore a scalar. Ex: 20 km

15. Displacement The change in position of an object

16. Difference b/n Distance & Displacement

17. Displacement = Change in position = Final position – Initial position

18.  x = x f – x i

19. Note: Displacement is not always equal to distance moved.

20. Displacement has a specific direction, therefore it is a vector.

21. Displacement can be positive (+) or negative (-).

22. On the x-axis displacement to the right is (+) and displacement to the left is (-). On the y-axis displacement upwards is (+) and displacement downwards is (-).

23. Speed Measure of how fast something is moving.

24. Is the distance covered per unit of time. ex: 72 km/hr or 20 m/s

25. Speed Units m/s, km/hr, or cm/s, same as velocity units. Since speed has no direction, it is a scalar.

26. The fast speed possible is the speed of light. Which is 3 x 10 8 m/s (299,792,458 m/s)

27. Instantaneous Speed Is the speed of an object at any instant.

28. Instantaneous Speed Is the speed of an object at any instant. Ex: speedometer reading

29. Average Speed The total distance divided by the time interval during which the displacement occurred. ( V avg )

30. Change in position V avg = ---------------------- Change in time total distance = ----------------------- time interval

31.  X v avg = ------  t

32.  x f – x i v avg = ---------  t

34. The cheetah averages 70 m/s over 30 seconds. How far does it travel in those 30 seconds?

35. Velocity Is the rate of change of displacement. It is speed with a direction. (vector)

36. The V avg can be (+) or (-), depending on the sign of the displacement.

37. Three Ways to Change Velocity

39. Ex 1: During a trip, a plane flies directly East with an average velocity of 35 m/s. What distance does the plane cover in 45 minutes?

40. G: V avg = 35 m/s,  t= 45 min= 2700 s U:  X = ? E:  X = ( V avg )(  t) S:  X=(35m/s)(2700s) S:  X = 94,500 m, E

41. With your group, work together to solve practice problems: 2, 4, and 6 on page 44 (HP). Do these problems in your notes. I will check for them in the notebook check. Also use the GUESS method.

42. 2) 3.1 km 4) 3 hr 6a) 6.4 Hour 6b) 77 km/hr South

43. Velocity is not the same as speed.

44. Uniform Motion Both velocity/speed and direction of the body/object remain the same.

45. Accelerated Motion Is when the velocity/speed of the object is changing.

46. Acceleration (a avg ) Is the rate of change of velocity. How fast you change your velocity

47. How do you know your accelerating?

49. a avg has direction and magnitude; therefore, it is a vector.

50. Change in Velocity a avg = --------------- Time interval for change

51.  v v f – v i a avg = --- = ---------  t  t

53. Units: m/s 2 or cm/s 2

54. If acceleration is negative (-) it means the object is slowing down or decelerating

55. Uniform Accelerated Motion Constant acceleration, meaning the velocity changes by the same amount each time interval.

56. Ex 2: A car slows down with an acceleration of –1.5 m/s 2. How long does it take for the car to stop from 15.0 m/s to 0.0 m/s?

57. G: a avg = -1.5 m/s 2, v f = 0.0m/s v i = 15 m/s U:  t = ? E: a avg = (v f – v i ) /  t or  t = (v f – v i ) / a avg

58.  t = (0m/s – 15 m/s) (-1.5 m/s 2 )  t = 10 s

59. Do practice problems, on page 49 (HP), #2 and 4. Work together with groups. These must be in notes and you need to use the GUESS Method to solve them.

60. Displacement (  x) depends upon: a avg, v i, and  t.

61. For an object moving with a uniform acceleration. The v avg is the average of the v i and the v f.

62. v i + v f v avg = ------- 2

63. By setting both v avg equations equal to each other.  x v i + v f ----- = -------  t 2

64. Multiply both side by  t  x = ½ (v i + v f )  t

65. Ex 3: Adam, in his AMC Pacer, is moving at a velocity of 27 m/s, he applies the brakes and comes to a stop in 5.5 seconds. How far did move before he came to a stop?

66. G: v i =27 m/s v f =0 m/s  t= 5.5 s U:  x E:  x = ½ (v i + v f )  t S:  x = ½(27 m/s+ 0m/s)5.5s S:  x = 74.3 m

67. Final velocity (v f ) depends upon: v i,  t, & a avg From the a avg equation:

68. Multiple both sides by time (  t) a avg  t = v f - v i

69. Then add v i to both sides. v f = v i + a avg  t

70. Ex 4: A pilot flying at 60 m/s opens the throttles to the engines, uniformly accelerating the jet at a rate of 0.75 m/s 2 for 8 seconds, what is his final speed?

71. G: a avg = 0.75 m/s 2,  t=8s, v i = 60 m/s U: v f = ? E: v f = v i + a avg  t S: v f =60 m/s + (0.75m/s 2 x 8 s) S: v f = 66 m/s

72. With our final speed equation, we can substitute it into the  x equation. This allows us to find  x without knowing v f.

73. We are going to substitute v f = v i + a avg  t into  x = ½ (v i + v f )  t

74. Where v f is, replace it with (v i + a avg  t)  x = ½ (v i + v i + a avg  t)  t  x = ½ (2v i  t + a avg  t 2 )  x = v i  t + ½ a avg  t 2

75. Ex 5: A plane flying at 80 m/s is uniformly accelerated at a rate of 2 m/s 2. What distance will it travel during a 10 second interval after acceleration begins?

76. G:  t= 10 s, a =2m/s 2, v i = 80 m/s, U:  x = ? E:  x = v i  t + ½ a  t 2 S:  x = (80 m/s)(10 s) + ½(2 m/s 2 )(10 s) 2 S:  x = 900 m

77. Open books to page 56 and read it. Try and follow the algebra and substitution for our last equation.

78. Final velocity after any displacement. v f 2 = v i 2 + 2a avg  x

79. Ex 6: A bullet leaves the barrel of a gun, 0.5 m long, with a muzzle velocity of 500 m/s. Find (a) its acceleration and (b) the time it was in the barrel.

80. G:  x=0.5 m, v f =500m/s, v i = 0 m/s U: a avg = ? E: v f 2 – v i 2 = 2a avg  x or a avg = (v f 2 – v i 2 ) / 2  x S: a avg =(500 2 – 0 2 )/(2 x 0.5) S: a avg = 2.5 x 10 5 m/s 2

81. G:  x =0.5 m, v f =500m/s v i = 0 m/s U:  t = ? E:  x = v i  t + ½ a avg  t 2 or  t 2 = 2  x /a avg S:  t 2 = 2(0.5) / 2.5 x 10 5 S:  t = 0.002 s

82. Free Falling Objects Galileo showed that a body falls with a constant acceleration of 9.81 m/s 2.

83. Acceleration of Gravity g = – 9.81 m/s 2 (For convenience we will use – 10 m/s 2 )

84. That means after 1 sec the object will have increased its speed by 10 m/s. So if starting from rest: After 1 sec – 10 m/s After 2 sec – 20 m/s After 3 sec – 30 m/s

85. Ex 7: A stone dropped from a cliff hits the ground 3 seconds later. Find (a) the speed with which the stone hits the ground, and (b) the distance it fell.

86. G:  t =3 sec,v i =0 m/s a avg = g = - 10 m/s 2 U: v f = ? E: v f = v i + a avg  t S:v f =0 m/s+(- 10 m/s 2 )(3s) S: v f = - 30 m/s

87. G:  t =3 sec, v i =0 m/s a avg = g = –9.81 m/s 2 U:  x= ? E:  x = v i  t + ½ a avg  t 2 S:  x = (0 m/s)(3 s) + ½ (– 10 m/s 2 )(3 s) 2 S:  x = – 45 m

88. Free falling bodies always have the same downward acceleration Even though an object may be moving upwards, its acceleration is downwards.

89. The velocity is positive, but is decreasing. When it reaches the peak, the velocity is zero, but still accelerating downwards.

90. Then the object begins to fall with a negative velocity.

91. A ball is thrown straight up with an initial velocity of 30 m/s.

92.  t (s)  y (m) V f (m/s)A avg (m/s 2 ) 0.00 1.00 2.00 3.00 4.00 5.00 6.00

93.  t (s)  y (m) V f (m/s)A avg (m/s 2 ) 0.00 - 10 1.00 - 10 2.00 - 10 3.00 - 10 4.00 - 10 5.00 - 10 6.00 - 10

94.  t (s)  y (m) V f (m/s)A avg (m/s 2 ) 0.0030 - 10 1.0020 - 10 2.0010 - 10 3.000 - 10 4.00 - 10 5.00 - 20 - 10 6.00-30 - 10

95.  t (s)  y (m) V f (m/s)A avg (m/s 2 ) 0.00030 - 10 1.002520 - 10 2.004010 - 10 3.00450 - 10 4.0040 - 10 5.0025 - 20 - 10 6.000-30 - 10

96. Ex 8: Amber hits a volleyball, so that it moves with an initial velocity of 6m/s straight upward. If the ball starts from 2 m off the floor. How long will it remain in the air before hitting the floor? Assume she is the last person to touch it.

97. G: V i = 6 m/s,  x = -2 m, a avg = -10 m/s 2 U:  t : There is no easy equation to use so we need to find  t

98. So we need to find the V f first. E: v f 2 = v i 2 + 2a avg  x S: v f 2 =(6m/s) 2 + 2(-10 m/s 2 )(- 2m) S: v f = +/- 8.7 m/s, since its moving downwards its – 8.7 m/s

99. Now we can find the time  t. E: a avg = (v f – v i ) /  t or  t = (v f – v i ) / a avg S:  t = (- 8.7 m/s – 6m/s) / (-10 m/s 2 ) S:  t = 1.47 s