2. Describing Motion KINEMATICS in One Dimension Chapter 2 “To understand motion is to understand nature.” Leonardo da Vinci

3. MECHANICS Study of motion, force and energy Kinematics How objects move Dynamics Why objects move

4. Kinematics Objectives ● Represent motion through the use of words, motion diagrams, graphs, and mathematical models. ● Use the terms position, distance, displacement, and time interval in a scientific manner to describe motion.

5. Reference Frames Any measurement of position, distance or speed must be made with respect to a frame of reference 80 km/h

6. Distance and Displacement Distance, d – total ground covered Displacement,  x change in position of an object (position, x, is measured from the origin of a chosen coordinate system) Position: x Scalar (has magnitude only) Vector (has magnitude AND direction) Unit [m]

7. Example - A car travels 400 km from Livingston to Philadelphia and then back 200 km to Trenton. What is the displacement of the car? What distance did the car travel? 0100200300400 kmX = LivingstonPhiladelphiaTrenton displacement distance 0 Distance and Displacement

8. Displacement Has magnitude (size) and direction. It is a VECTOR 123 x(m) Vectors are represented by arrows

9. Displacement Has magnitude and direction. It is a VECTOR 123 x(m)

10. Average Speed and Average Velocity Average speed (s) describes how fast a particle is moving. It is calculated by: Average velocity (v) describes how fast the position is changing with respect to time: always positive sign gives direction in 1 Dimension Scalar (has magnitude only) Vector (has magnitude and direction) Average speed and average velocity often have the same magnitude, but not always Unit [m/s]

11. Example - A car travels 400 km from Philadelphia to Livingston in 2 hours and then back 200 km to Trenton in 1 hr. What is the car’s speed and velocity? 0100200300400 kmX = LivingstonPhiladelphiaTrenton Average speed Average velocity xx

12. Example The position-time graph shows the progress of two runners, A and B. a) When does runner B pass runner A? b) Where does runner B pass runner A? c) What is the starting position for runner A? runner B? d) After 10 hrs, what is the average velocity of runner A? After 10 hrs, what is the average velocity of runner B? e) If the finish line is at 40 km, who won the race?

13. Graphical Representation of Motion t x A B xx tt slope Steepness = speed Sign = direction Velocity = speed + direction Position-Time Graph slope

14. Average Velocity from a Graph t x A B xx tt Mathematical Model x = position x 0 = initial position v av = average velocity t = time

15. Graphs of Motion Mathematical Model x 0 = 20 m v av = 2 m/s Slope = Mathematical Model =Area Area =  x = 20m  x =20m SLOPE AREA  x =20m (init and final positions unknown. ONLY KNOW DISPLACEMENT) (UNIFORM VELOCITY)

16. What is happening in this graph? -10-505101520 mX = t=0 s START End t=6 s v = 1s 2s3s4s5s Mathematical Diagrammatic Graphical

17. Plot the corresponding v-t graph x 0 = 20 m v av = -5 m/s Slope SLOPE

18. Draw the corresponding v-t graph v-t SLOPE

19. Draw the corresponding x-t graph Area=  x -6m  x= +12m 3m  x= -16m AREA v-t

20. Average Velocity and Instantaneous Velocity AVERAGE VELOCITY: -40-20020406080 kmX = -40-20020406080 kmX = Start t=0End t=2 hr Average velocity only depends on the initial and final positions. These 2 cars have the same average velocities but different velocities at each instant. When the velocity is not uniform, the instantaneous velocity is not the same as the average velocity.

21. Average Velocity Remember that graphically the average velocity between the A and B is the slope of the line connecting them, or the slope of the secant line. t x A B xx tt Mathematical Graphical

22. x A B t What happens if A and B become closer to each other? How can we find the instantaneous velocity at a point?

23. What happens if A and B become closer to each other? x A B t Average Velocity from a Graph

24. x A B t What happens if A and B become closer to each other? Average Velocity from a Graph

25. The line “connecting” A and B is a tangent line to the curve. The velocity at that instant of time, v, is represented by the slope of this tangent line. A and B are effectively the same point. The time difference is effectively zero. x A B t Instantaneous Velocity from a Graph

26. The instantaneous velocity at B is the slope of the tangent line at B x B t Instantaneous Velocity Mathematical Graphical

27. Acceleration Average acceleration describes how the velocity changes with respect to time. It is calculated by: Vector Instantaneous acceleration describes how the velocity changes over a very short time interval: Acceleration tells us how fast the velocity changes whereas velocity tells us how fast the position changes. SI units: m/s 2

28. Example - A car accelerates along a straight road from rest to 24 m/s in 6.0 s. What is the average acceleration? Average acceleration START v 1s2s3s4s5s a x 481216 4m/s 2 20m/s 0 0 How far did the car travel in 6s?

29. START 1s2s3s4s5s x t=0 t=5 t=4 t=3 t=2 t=1 x Position-Time Graph v (+) a (+)  v>0 a >0

30. START v 1s2s3s4s5s a x START v 1s2s3s4s5s a x Speeding up in + direction Slowing down in + direction a and v SAME direction a and v OPP direction 0 0

31. START v 4s3s2s1s0 a x Slowing up in - direction a and v OPP direction 5s START v 4s3s2s1s 0 a x Speeding up in - direction a and v SAME direction 5s

32. Displacement and velocity are in the direction of motion When acceleration is in the SAME direction as velocity, the object is speeding up When acceleration is in the OPPOSITE direction to velocity, the object is slowing down

33. START 4s3s2s1s5s x t=5 t=0 t=2 t=3 t=4 t=1 v (-) a (+) Slowing down in - dire ct ion What is this object doing? 0  v>0 a >0

34. START 1s2s3s4s5s x t=0 t=5 t=4 t=3 t=2 t=1 x v (+) a (+) Position-Time Graph Speeding up in + direction 0  v>0 a >0

35. START 1s2s3s4s5s x t=0 t=5 t=3 t=2 t=1 t=4 x v (+) a (-) Position-Time Graph Slowing down in + direction 0  v<0 a <0

36. t=5 t=0 t=1 t=2 t=3 t=4 Speeding up in - direction START 4s3s2s1s5s x 0 v (-) a (-)  v<0 a <0

37. START 4s3s2s1s5s x t=5 t=0 t=2 t=3 t=4 t=1 Slowing down in - dire ct ion 0 v (-) a (+)  v>0 a >0

38. Draw the corresponding v-t and a-t graphs x-t graph v-t graph a-t graph Checks: v – a+ Object is slowing down in - direction

39. t=5 t=0 t=1 t=2 t=3 t=4 x va Speeding up in - direction START 4s3s2s1s5s x 0 What is this object doing?

40. AVERAGE VELOCITY SLOPE = For motion with constant velocity AREA under curve RUNNING TOTAL Graphical Representation of Motion (Uniform Velocity)  x =3  x =6  x =9  x =12  x =15 SLOPE AVERAGE acceleration SLOPE = AREA NO acceleration Connect with straight line

41. AVR Velocity SLOPE = AREA RUNNING TOTAL  v =-9 slope AVR acceleration SLOPE = AREA No physical meaning slope AREA SLOPE = acceleration Connect with curved line Constant Acceleration Motion  x =13.5  x =-6 AVR jerk

42. SLOPE x-t SLOPE v-t AREA a-t AREA v-t PHYSICS DEPARTMENT STORE

43. A car starting from rest accelerates at a constant rate of 1 m/s 2 for 3 s and then cruises at constant velocity for 2 sec. What is the displacement from 0 s to 5.0 s  x = 4.5 + 6 = 10.5 m Area under v-t curve

44. Construct the corresponding x-t and a-t curves AREA Curved (acceleration) Straight (constant v) slope

45. Construct the corresponding v-t and a-t curves slope A B C A B C

46. Estimate the displacement from 0 s to 4.0 s  x = 2 -2 = 0 m Area under v-t curve

47. AREA Construct the corresponding x-t and a-t curves All Curved (acceleration) slope

48. A)What is the acceleration at 1 s, at 3 s and at 6 sec? B) What is the total displacement for the entire trip? C)What is the distance traveled for the trip?  x 0-10 = 6m a 1 =2m/s 2, a 3 =0 m/s 2, a 6 =-4 m/s 2 d 0-10 = 30m

49. Construct the corresponding x-t and a-t curves curved straight slope AREA A B C D E F

50. Example - A car starts with initial velocity of 5 m/s and accelerates for 4.0s along a straight road at a constant acceleration of 2 m/s 2. After 4 s, how far has the car traveled?

51. Do Now (solve graphically): A car moving at 12 m/s is 36 m away from a stop sign. What acceleration will stop the car exactly at the stop sign? a = -2 m/s 2 t stop Area  x=36m Slope=a 12 v-t

52. = SLOPE x-t rise run xx tt v av = KINEMATIC EQUATIONS (Constant Acceleration) = rise run vv tt a av = a = constant acceleration Initial velocity Amount velocity changes due to accel (  v) Equation of v-t graph SLOPE v-t

53. KINEMATIC EQUATIONS (Constant Acceleration)  x = AREA v-t = area( ) + area( ) Displacement if there was no acceleration How much further travel with acceleration vv

54. KINEMATIC EQUATIONS (Constant Acceleration)  x = AREA = area( ) + area( ) (For linear velocity or constant a)

55. Uniform Accelerated Motion Equation representing x-t graph Equation representing v-t graph Equation representing a-t graph Mathematical Graphical

56. KINEMATIC EQUATIONS (Constant Acceleration) (definition of average velocity) (average velocity for constant acceleration) (definition of avr a)  (time independent)

57. KINEMATIC EQUATIONS (ONLY for Constant Acceleration) 1. 3. 4. 5. 2. (If a = 0) v 0 =v=v av

58. Example - A car starts with initial velocity of 5 m/s and accelerates for 4.0s along a straight road at a constant acceleration of 2 m/s 2. After 4 s, how far has the car traveled? v 0 = 5 m/s v a = 2 m/s 2  x = ?? m t = 4s

59. Example: A car moving at 12 m/s is 36 m away from a stop sign. What acceleration will stop the car exactly at the stop sign? v 0 = 12 m/s v = 0 a = ? m/s 2  x = 36 m t a = -2 m/s 2

60. KINEMATICS PROBLEMS A passenger jet lands on a runway with a velocity of 70 m/s. Once it touches down, it accelerates at a constant rate of -3 m/s 2. How far does the plane travel down the runway before its velocity is decreased to 2 m/s, its taxi speed to the landing gate? v 0 = 70 m/s v = 2 a = -3 m/s 2  x = ? m t  x = 816 m xx vivi a vfvf a

61. KINEMATICS PROBLEMS A runner goes 12 m in 3 s at a constant acceleration of 1.5 m/s 2. What is her velocity at the end of the 12 m? v 0 = v = ? m/s a = 1.5 m/s 2  x = 12 m t = 3 s v 0 = 1.75 m/s v = 6.25 m/s

62. The U.S. and South Korean soccer teams are playing in the first round of the world cup. An American kicks the ball giving it an initial velocity of 4.0 m/s. The ball rolls a distance of 12.0 m and is then intercepted by a South Korean player. If the ball accelerated at -0.50 m/s 2 while rolling across the grass, find its velocity at the time of interception. KINEMATICS PROBLEMS v 0 = 4.0 m/s v = ? m/s a = -0.50 m/s 2  x = 12.0 m t = v = 2.0 m/s

63. A drag racer starting from rest, speeds up for 402 m with an acceleration of 17.0 m/s 2. A parachute then opens, slowing the car down with an acceleration of -6.10 m/s 2. How fast is the racer moving 350 m after the parachute opens? v 01 = 0 v 1 = a 1 = 17.0 m/s 2  x 1 = 402 m t 1 = Speeding up (a 1 ) Parachute braking (a 2 ) v 02 = v 1 = v 2 = ? a 2 = -6.10 m/s 2  x 2 = 350 m t 2 = v 1 = 116.9 m/s 116.9 m/s v 2 = 96.9 m/s

64. Just as you turn onto the main avenue from a side street with a stop sign, a city bus going 30 mph (13.4 m/s) passes you in the adjacent lane. You want to get ahead of the bus before the next stoplight which is two blocks away. Each block is 61m long and the side streets are 7.6 m wide. If you increase your speed at a rate of 2.0m/s each second, will you make it t B = 9.7 s to stoplight t C = 11.4 s to stoplight NO, you will not make it BUS v B = 13.4 m/s  x = 129.6 m t B = ? s CAR v 0 = 0 v = a = 2.0 m/s 2  x = 129.6 m t C = ? s

65. Car chase problem: Just as you turn your car onto the main avenue from a side street with a stop sign, a city bus going 30 mph (13.4 m/s) passes you in the adjacent lane. If you increase your speed at a rate of 2.0m/s each second, when will the car catch the bus? BUS (a = 0) v B = 13.4 m/s  x B =  x C =  x t B = t C = t = ? CAR v 0 = 0 v = a = 2.0 m/s 2  x B =  x C =  x t C = t B = t = ? Chase problem: If both vehicles go the same distance in the same time, they must have the same v av

66. Car chase: A car is stopped at a traffic light. When the light turns green at t=0, the car speeds up with constant acceleration. At the same instant, a city bus going at a constant speed, v B, passes the car. The car catches up to the bus at x=D. Sketch the bus and car v-t plots on the same graph. Label the time, t D, when the car catches up to the bus on the graph. Use the graph to a)find the speed of the car when it catches up to the bus (hint: when the car catches up to bus, at t D, the areas under the v-t curves are equal) b)derive an expression for the acceleration of the car in terms of V B and D. vBvB 2v B tDtD v C = 2v B Slope v-t D is v-t Area from 0-t D At t D http://tube.geogebra.org/student/m161682 tDtD X

67. Car chase: A car is stopped at a traffic light. When the light turns green at t=0, the car speeds up with constant acceleration. At the same instant, a city bus going at a constant speed, v B, passes the car. The car catches up to the bus at x=D. a)Sketch the bus and car x-t plots on the same graph. Label the position (D) and the time (t D ) when the car catches up to the bus on the graph. b)By setting the car and bus positions equal, derive an expression for the acceleration of the car in terms of V B and D. tDtD D At (t D, D) Need to put t D in terms of v B and D

68. Reaction Time Problem: Estimate the minimum stopping distances for a car, which are important for traffic safety and traffic design. The stopping distance depends on the reaction time of the driver (a = 0), initial and final speeds of the car and acceleration of the car when the driver starts to apply the brakes (a ≠ 0). For a dry road and good tires, good brakes can decelerate a car at a rate of about 5-8 m/s 2. Calculate the total stopping distance for an initial velocity of 100 km/hr (28m/s≈62mph) and assume the acceleration of the car is -6.0 m/s 2. Reaction time for normal drivers varies from 0.3- 1.0 sec; take it to be 0.50s for this problem. v 0 =v = 28m/s  x 1 = ? t 1 = 0.5 s  x =  x 1 +  x 2 =79.3 m v 0 = 28 m/s v = 0 a = -6 m/s 2  x 2 = ? m t 2 = Reaction time (a=0) Braking

69. Two spacecraft are 13,500 m apart and moving directly toward each other. The first spacecraft has a velocity 540 m/s and accelerates at a constant -15 m/s 2. They want to dock, which means that they have to arrive at the same position at the same time with zero velocity. A)What should the initial velocity of the second spacecraft be? B)What should be its constant acceleration? KINEMATICS PROBLEMS Spacecraft 1 v 0 = 540 m/s v = 0 a 1 = -15 m/s 2  x 1 = ? m t 1 = ? s goes  x 1 = 9720 m in t 1 = 36 s Spacecraft 2 v 0 = ? m/s v = 0 a 2 = ? m/s 2  x 2 =  x 1 – 13,500 = -3780 m t 2 = t 1 = 36 s v 2 = -210 m/s a 2 = 5.83 m/s 2

70. Free Fall Chapter 2 Baumgarten FreeFall

71. Aristotle (384 BC – 322 BC) - Greek Philosopher - student of Plato - teacher of Alexander the Great Aristotle’s views on the physical sciences had profound influence for 2000 years. They were ultimately replaced by Newtonian mechanics. Each of the four earthly elements has its natural place; the earth at the centre of the universe, then water, then air, then fire. When they are out of their natural place they have natural motion, requiring no external cause, which is towards that place; so bodies sink in water, air bubbles rise up, rain falls, flame rises in air.

72. FREE FALL

73. Galileo Galilei (1564 – 1642) Italian physicist, mathematician, astronomer and philosopher father of modern science Galileo proposed that a falling body would fall with a uniform acceleration, as long as the resistance of the medium through which it was falling was negligible (in vacuum).

74. http://www.teachersdomain.org/asset/phy03_ vid_galileoplane/ All unsupported objects fall towards the earth with the same acceleration. How did Galileo prove this?? NO watches/clocks NO video/time-lapse ….. Free Fall He slowed down the motion with inclined planes

75. All unsupported objects fall towards the earth with the same acceleration. We call this acceleration the "acceleration due to gravity" and it is denoted by g. g = 9.8 m/s 2 Keep in mind, ALL objects accelerate towards the earth at the same rate. g is a constant! Free Fall is motion with CONSTANT ACCELERATION FREE FALL In vacuum (no air resistance )

76. http://www.youtube.com/watch?v=WOvwwO-l4ps&feature=related Apollo mission to the moon http://www.youtube.com/watch?v=KDp1tiU sZw8

77. Acceleration Due to Gravity g = -9.8 m/s 2 Velocity (m/s) Way Down Velocity is increasingly negative Speed is increasing SPEEDING UP (v-, a-) DROP When an object is dropped, the initial velocity is zero. + -

78. + - THROWN UP When an object is thrown, it has a nonzero initial velocity. Max Height Velocity is ZERO Acceleration = -9.8 m/s 2 Way Down Velocity gets more negative Speed is increasing SPEEDS UP (v-, a-) Acceleration = -9.8 m/s 2 Way Up Velocity is decreasing Speed is decreasing SLOWS DOWN (v+, a-) Acceleration = -9.8 m/s 2 Acceleration Due to Gravity g = -9.8 m/s 2 Velocity (m/s)

80. An object is thrown upward with initial velocity, v o + - THROWN UP v0v0 v = -v 0 At top, v = 0 a = -9.8 What is the velocity right before it lands at the same height?

81. Maximum Height What is the velocity of the ball at the top of its flight? 0 m/s What is the acceleration of the ball at the top of its flight? -9.8 m/s 2 What would happen if the both velocity and acceleration are zero at the top of the ball’s flight? The ball would float.

82. Kinematics Equations for uniformly accelerated motion Horizontal (a x =0) Vertical (FREE FALL) For FREE FALL:

83. Problem Solving for Free Fall (Points to Remember) Acceleration is always -9.8 m/s 2 At maximum height – Velocity is zero – Acceleration is -9.8 m/s 2 When an object is dropped, initial velocity is zero When an object is thrown into the air, it must have an initial velocity Time up is equal to time down to initial height At any given height, velocity up = –(velocity down).

84. DO NOW A rocket is fired straight up from the ground. It returns to the ground 10 seconds later. What was its launch speed? v i = ? m/s v f = 0 m/s a = -9.8 m/s 2 ~ 10 m/s 2  y = t = 5 s v i = -50 m/s v 0 = ? m/s v = -v 0 a = -9.8 m/s 2 ~ 10 m/s 2  y = t = 10 s

85. An arrow is fired into the air and reaches its highest point 3 seconds later. What was its velocity when it was fired? v 0 = ? m/s v = 0 m/s a = -9.8 m/s 2 ~ 10 m/s 2  y = t = 3 s v 0 = 30 m/s

86. An acorn falls from an oak tree. You note that it takes 2.5 seconds to hit the ground. How fast was it going when it hit the ground? v i = 0 m/s v f = ? m/s a = -9.8 m/s 2  y = t = 2.5 s v f = -24.5 m/s

87. Example: A dropped ball falls 8.0 m. What is its final velocity? SIGNS!! v 0 = 0 m/s v = ? m/s a = -9.8 m/s 2  y = -8.0 m t = v = -12.6 m/s y i = 0 y f = -8

88. A rock is thrown straight upwards with a speed of 15 m/s from the edge of a cliff 50 m high. The rock reaches a maximum height of above the edge and then falls down to the bottom of the 50 m cliff. Use g = 10m/s 2 A)How much later does the rock reach the bottom of the cliff? B)What is the impact velocity (velocity right before landing) ? C)What is the maximum height of the rock? t = 5s v0v0 50m v = -35m/s  y max = 11.25m

89. DO NOW What is the initial velocity of this object? v i = ? m/s v f = 0 m/s a = -9.8 m/s 2 ~ -10 m/s 2  y = t = 4 s v i = 40 m/s

90. Example: A wrench falls from a helicopter which is rising steadily at 6.0 m/s a)What is the velocity of the wrench 2.0 s later? b)How far below the helicopter is the wrench? Wrench v 0 = 6.0 m/s v = ? m/s a = -9.8 m/s 2  y = t = 2.0 s v = -13.6 m/s Helicopter v = 6.0 m/s  y = t = 2.0 s  y = 12 m  y = -7.6 m So the wrench is 19.6 m below the helicopter. y = 0 y = -7.6 y = 12

91. Relationships (cont.) An arrow is launched straight up at speed v 0 from ground level. Determine how much each of the following changes if the launch speed is tripled. a.max height a.hang time a.impact speed 3x 9 x 3x Answers Equation relating V 0 and  y max (constants: a y, v=0) Equation relating V 0 and t (constants: a y,  y=0, v=-v 0 ) Relationships

92. Let’s use the kinematics equations to answer these: 1. A mango is dropped from a height h. a. If dropped from a height of 2 h, would the impact speed double? b.Would the air time double when dropped from a height of 2 h ? 2.A mango is thrown down at a speed v. a.If thrown down at 2 v from the same height, would the impact speed double? b.Would the air time double in this case? no, increase by  2 x

93. To determine how high a cliff is, a llama farmer drops a rock, and then 0.800 s later, throws another rock straight down at a velocity of −10.0 m/s. Both rocks land at the same time. How high is the cliff? Rock 1 v 0 = 0 m/s v f = a = -9.8 m/s 2  y = ? t 1 = t  y = -24.8 m Rock 2 v 0 = -10 m/s v f = a = -9.8 m/s 2  y = ? t 2 = t – 0.8

94. Wiley in Free Fall coyote and anvil What is wrong with the physics? How far did Wile E. Coyote fall?