THE KINEMATICS (GRADE 12 STEM STRAND).pptx

THE KINEMATICS (GRADE 12 STEM STRAND).pptx

1. KINEMATICS IN MOTION

2. LEARNING TARGETS Convert a verbal description of physical situation involving uniform acceleration in one dimension into a mathematical description Interpret displacement and velocity, respectively as areas under velocity vs time and acceleration vs time curves interpret velocity and acceleration, respectively as slopes of position vs time and velocity vs time curve

3. LEARNING TARGETS Construct velocity vs time and acceleration vs time graphs respectively corresponding to a given position vs time-graph and velocity vice versa Solve for unknown quantities in equations involving one dimensional uniformly accelerated motion, including free fall motion Solve problems involving one dimensional motion with constant acceleration in contexts such as but not limited to the tail gating phenomenon, pursuit rocket lunch and free-fall problems

4. MOTION It is one of the most common physical phenomena.

5. MOTION . we see motion in the activities people do every day

6. Walking Jogging Running Riding a car to go to school

7. MOTION . Motion can also be observed in nature

8. Clouds moving Raindrops falling Wind blowing the leaves of a tree Water moving in an ever ending cycle

9. The study of motion is divided into kinematics and dynamics.

10. Kinematics – is a quantitative description of motion without reference to its physical clauses, here we define terms, like displacement, velocity and acceleration

11. Dynamics is the study of the relationship between motion and force.

12. Our first step in the study of motion is to define concepts for motion in a straight line. Translation is the physical term for straight-line motion, We then extend our discussion to projectile motion.

13. FRAME REFERENCE AND MOTION The term position refers to the location of an object with respect to some reference frame. A reference frame is a physical entity such as ground, room, or a building to which position of an object is being referred. For instance, we say that the gymnasium is 20m to the right of the main gate.

14. When an object is undergoing a continuous change in position, we say that the object is moving. Motion is a relative term. It depends on the reference frame where motion is being observed.

15. Imagine that you are riding in a car with another passenger. Observing your fellow passenger reveals no change in position. He is at rest as far as you are concerned. However, to an observer standing by the road, you and the passenger are in motion. In everyday life, we always refer motion relative to the earth.

16. DRIVERS NO MORE

17. Youtube : https://you.tube/tnmJZ3rCKeU

18. What are the 10 most common causes of road accidents

19. According to the video, the 10 most common causes of road accidents mentioned in this link: Bad overtaking Avoiding hitting another vehicle Avoiding hitting a pedestrian Mechanical malfunction Bad turning Lost brakes Inattentively moving backwards Lost control Drunk driving Drive error

20. One of the leading causes of vehicular accidents is driver error, which is often the result of distracted driving. The most common form of distracted driving is the use of cellphone by either calling or texting.

21. Studies have shown that the human brain cannot multitask, which means it cannot perform multiple tasks at time. Researchers have found that the brain can only “micro-task,” which involves handling only one task at a time and switching between tasks very quickly.

22. Other causes of driver error include intoxication and fatigue. An intoxicated or fatigued person has poor vision, decreased reaction time, and a tendency to fall asleep while driving. A fatigued driver is likely to miss traffic control signals compared to an alert driver.

23. Can we do away with the driver?

24. The answer is yes! It seems like the driverless or self-driving car is well on its way to becoming a reality.

25. SkyDrive Inc. has begun sales of its flying vehicle to individuals, the startup based in Toyota, Aichi Prefecture, announced Thursday. The electric vertical takeoff and landing aircraft costs $1.5 million , or about ¥200 million. Deliveries will start in 2025 or later.

26. DISTANCE AND DISPLACEMENT

27. Distance A scalar quantity that refers to the total length of path taken by an object in moving from its initial position to final position Or d refers to the actual length of path taken by an object in moving from its initial position to its final position.

28. Displacement a vector quantity that refers to the straight –line distance between an object’s initial and final positions, with direction toward the final position Or d refers to the straight line distance between its initial position and final position, with direction toward the final position.

29. SAMPLE PROBLEM Suppose that in going to school, you walked 40 m east and 30 m north.(a) What was the total distance that you walked? (b) What was your total displacement? When you arrived in school, you found out that classes were suspended because of bad weather. You went back home following the same path.(c) What was the total distance you walked in going to school and back home (d) what was your total displacement

30. a. The total distance that you walk is 30m + 40m = 70 m

31. b. Using the Pythagorean theorem d = √(30m) 2 + (40m) 2 = 50m solving for direction Θtan -1 {30m/40m} = 37 o N of E Thus your displacement is 50m, 37 N of E

32. c. The total distance is 30m + 40m+ 30m+ 40m = 140m

33. d. Since you are back to where you started, your displacement is zero

34. PRACTICE EXERCISE A student walk 50m due east. Realizing that he forgets his book, he decided to go back to his house. He had traveled 35m due west when he meets his brother, he turn east and walks his way to school an additional displacement of 50m. Find (a) total distance the student walked and (b) his resultant displacement

35. Speed and Velocity

36. The term “speed” and “velocity” are used interchangeably in everyday situation. However in physcis they have distinct meaning

37. Speed refers to how fast an object is moving or the distance that a body moves un a unit time. It is a scalar quantity

38. Velocity refers to the art of change of position When the speed of a body is associated with a direction, the result is the velocity of the body. A vector quantity

39. Velocity The speed of the body is the magnitude of its velocity. The SI unit for speed and velocity is meter per second, m/s.

40. An object normally changes its speed while moving. Hence, it is necessary to distinguish between average speed and instantaneous speed. The average speed of a body is the total distance traveled divided by the time spent in travelling the total distance average speed = total distance total time v = Δ d Δ t

41. Δ t and Δ d are time and length intervals respectively. Δ t means t-t o where t and t o are final and initial times, respectively.If t = 0 then Δ t = t. Similarly Δ d = d – d o where d is the final position and d o is the initial position. Taking the initial position as zero, then Δ d = d. v = d/t

42. Sample Problem Every morning, you jog around a 250 m track four times in 30 minutes. What is your (a) average speed and (average velocity)

43. Solution a. The total distance you have jogged is 4 x 250 m = 1000m, thus your average is v =d/t = 1000m/1800s v = 0.6m/s

44. b. You have no resultant displacement since you are back to where you started. Therefore, your average velocity is zero.

45. Practice Exercise A student drives 5.5 km from his home to school in 35 minutes but makes the return trip in only 25 minutes. What are the average speeds and average velocities for (a ) each half of the round trip and (b) total trip?

46. a. average speed =total distance/total time =5.5km /35 mins = o.16km/ mins = 5.5 km/25 mins = 0.22 km/ mins b. average velocity = total displacement/total time d = 5.5 x 2 = 11km t = 35 mins +25 mins = 60mins=1hr v = d/t = 11km/ 1hr = 11km/ hr

47. Some common units of speed are kilometres per hour(km/ hr )

48. In the real world, objects move with changing velocities. A new concept called acceleration has been defined to describe how velocity changes.

49. ACCELERATION Is change in velocity with respect to time. Velocity can change in three ways Change in speed either increase or decrease Change in direction Change in speed as well in direction

50. Thus an object is said to be accelerated when it is either moving with changing speed, moving with constant speed but with changing direction, or moving with changing speed as well as changing direction.

51. Acceleration = change in velocity time a = Δ v Δ t Since acceleration is defined in terms of velocity, it is also vector quantity. The SI unit for acceleration is meter per second, m/s/s or ms 2

52. Sample problem In which cases is the car accelerating A car changing lanes at constant speed A car speeding in an effort to beat the red light A car slowing down while making a left turn

53. ANSWERS In all three cases, the car is accelerating. In case A, there is acceleration because of a change in direction. The car is accelerating in case b because of the change in speed. In case C, the car accelerates because it changes its speed and direction.

54. PRACTICE EXERCISE Suppose you are caught in a heavy traffic in Mabuhay Rotanda . Are you accelerating when you are going around Rotanda at constant speed

55. MOTION IN STRAIGHT LINE In order to better describe the motion of a body, we shall be using a Cartesian coordinate system as our frame of reference. We shall consider the origin as the start of motion. For one-dimensional motion, we shall consider motion along x – axis.

56. MOTION IN STRAIGHT LINE Position to the right of the origin shall be considered positive. Position to the left of the origin shall considered negative. A positively velocity would indicate motion to the right of the origin. A positive velocity would indicate motion to the right of the origin. A negative velocity would mean motion to the left of the origin.

57. UNIFORM MOTION

58. The simplest type of motion is uniform motion. For uniform motion, the velocity is constant, the acceleration is zero, and the instantaneous velocity is equal to the average velocity. Its displacement d x may be obtained by multiplying its constant velocity v by the time t.

59. SAMPLE PROBLEM A car moving at constant speed travels 30m in 5s. What is the speed of the car? How far will the object move in 10s?

60. Solution We are given that in 5s, the distance traveled at constant speed is 30m. The relationship d x = v t is appropriate. Substituting the given values 30m = v (5s) v = 6m/s b. Using again the relation d x = v t d x = (6m/s) (10s) 60 m

61. Practice exercise Two cyclist race against the clock in a 50.0km cross-country route. Cyclist A travels at a constant speed of 50.0km/hr. Cyclist B started 15.0 min late but manages to catch up with a cyclist A at the finish line. What is the speed of the cyclist B assuming that his speed is constant?

62. V = d/t =50.0km/.75 = 67km/ hr

63. UNIFORMLY ACCELERATED MOTION A more common type of motion would be that in which velocity is changing at a constant rate. Since the discussion is limited to straight line motion, acceleration here would mean change in speed rather than change in direction.

64. UNIFORMLY ACCELERATED MOTION If an object changes its velocity from an initial velocity v o and a final velocity v f , during a time interval t , its acceleration is given by a= v f -v o / t

65. Sample Problem A Nissan Sentra is stopped at a traffic light. When the light turns green, the driver accelerates so that the car’s speedometer reads 10m/s after 5s. What is the car’s acceleration assuming its constant? Solution We are given that v o =o and v f = 10m/s at t = 5s. Substituting these values in the equation for acceleration, a=10m/s – 0/5 = 2m/s/s = 2ms 2

66. A particle is moving with uniform acceleration in a straight line was first observed to be moving at 4.5m/s. After 10s, it was moving at 6.8m/s. Find its acceleration.

67. Knowing the initial velocity , the acceleration , and the time during which the object is accelerating, the previous equation may be rearranged to arrive at an equation for final velocity. v f = v o + a t

68. The displacement of a uniformly accelerating object may be obtained from the definition of average velocity. v = d x /t but v = ½( v f + v o ) solving for d x d x =1/2( v f + v o ) t Replacing v f by v o + a t and simplifying, d x =1/2( v o + a t + v o ) t d x + v o t = ½ a t 2

69. Sample Problem A racer accelerates from rest at a constant rate of 2.0m/s/s. (a) How fast will the racer be going to the end of 6.0 s? (b) How far the racer traveled during this time? Solution: Given v o = 0 a = 2.0m/s/s t = 6.0 s v f =? d x = ? Using the equation for final velocity and considering only the scalar values, v f = v o + a t v f = 0 +(2.0m/s/s)(6.0 s) v f = 12 m/s b. Substituting the given values in the equation d x = v o t +1/2 a t 2 d x= ½(2.0m/s/s) (6.0 s) 2 d x = 36m

70. Another useful equation relating final velocity, initial velocity, and acceleration may be obtained. Starting from d x = v t and v = ½( v f + v o ) we obtain d x =1/2( v f + v o )t Replacing t by v f - v o / a d x = ½( v f + v ) { v f - v o / a } Performing the operation and simplifying 2ad x = v f 2 – v o 2

71. PRACTICE EXERCISE A car starting from the rest passes two successive milestones A and B that are 250m apart. It takes 60.0 s to pass A and another 60.0 s to pass A and B. Find the velocities of the car when it passes milestone A and milestone B.

72. Assume that the acceleration is constant from start to milestone A: V A = at = 60 a From milestone A to milestone B d x = v o t + 1/2at 2 25o = v o t + 1/2at 2

73. 25o = v o t + 1/2at 2 250 m = 60s(a) 60s + ½ a (60s) 2 250 = (3600s) a + 3600/2 a 250 = (3600s) a + (1800s) a 250 m= 5400 a a = 250m/5400s 2 a = 0.o46m/s 2

74. Velocities of each car V A = (0.046m/s2)(60s) = 2.8 m/s or 3m/s V B = V A + at = 2.8m/s+ (0.046m/s 2 )(60s) = 5.56m/s or 6m/s

75. Sample problem A car has uniformly accelerated from rest to a speed of 25m/s after travelling 75m. What is its acceleration? Solution Given: v o = o, v f =25m/s, and d x = 75m Substituting the given values in the equation 2ad x = v f 2 – v o 2 2a(75 m) = (25m/s) 2 – 0 = 4.2m/s 2

76. 2a(75 m) = (25m/s) 2 – = 150 a = 625 – 0 a = 625/150 a = 4.2m/s 2

77. PRACTICE EXERCISE A driver travelling at 12m/s sees a stalled vehicle 17m ahead of his car. He steps on the brakes to slow down the car. At what rate must the car slow down so that it stop just before hitting the stalled vehicle? Assume that the driver’s reaction time is .25s. Reaction time is the time for a person to notice, think, and act in response to a given situation.

78. 2ad x = v f 2 – v o 2 2a(17m-12m/s)(0.25s) = 0-12m/s 2 a = - 5.1m/s 2

79. FREE FALL

80. Perhaps you observed that what goes up always come down. In the absence of air resistance, it is found that all bodies regardless of size and weight at the same location above the earth’s surface fall vertically with the same acceleration.

81. This very idealized motion in which air resistance is neglected and the acceleration is constant is called free fall. The acceleration of a freely falling body is called acceleration due to gravity. This is denoted by a letter “g” and is equal to - 9.8m/s 2 at the surface of the earth. This acceleration is downward and directed toward the center of the earth.

82. In r with r eality , g decreases with increasing altitude according to the equation g h = g R 2 /(R + h) 2 where R is the radius of the earth = 6.37 x 10 6 m, h is the altitude, and g = -9.8ms 2 , g h is the acceleration due to gravity at an altitude h . However if h is very small compared to R then g h = g . For our purposes, we shall assume that h is negligible and g = -9.8ms 2

83. We may threat free fall as a case of uniformly accelerated motion. All the equations that we have derived may be used to analyze free fall. Since motion is along the vertical direction, it is better to replace d x with d y in our equations and a with g . Correspondingly we have d y = v ot =1/2 g t 2 2 gd y =v f 2 –v o 2 g = v f – v o /t

84. The following signs conventions shall also used: a. Distances above the origin are positive, while distances below the origin are negative. b. Upward velocities are positive, downward velocities are negative. c. g is always negative The origin is the point of release.

85. The notion of the object thrown upward and eventually to its starting point exhibits two symmetries: time symmetry and speed symmetry. Time symmetry means that the time required for the object to reach maximum height equals the time for it to return from its maximum height to its starting point.

86. Speed symmetry show that at any displacement above the point of release, the speed of the body during the upward trip equals the speed during the downward trip.

87. Sample Problem A boy tosses a coin upward with a velocity of +14.7m/s. Find (a) the maximum height reached by the coin, (b) time of flight and (c) velocity when the coin returns to the hand (d) Suppose the boy failed to catch the coin and the coin goes to the ground, with what velocity will it strike the ground? The boy’s hand is 0.49 m above the ground .

88. Solution We are given that the initial velocity is +14.7m/s. Point A is the point of release. Let the origin be at the point A a . Distance AB is the maximum height reached by the coin from the point of release. At point B, final velocity = o. Representing distance AB by dymax

89. Solving for d ymax 2 g d ymax =v f 2 –v o 2 2(-9.8m/s 2 ) d ymax = 0 – (+14.7ms) 2 d ymax = 11.0

90. b.Consider portion AB in solving for time to go from point A to point B. g = v f – v o / t -9.8ms 2 = 0 –(+14.7m/s)/t t = 1.50s

91. c. Starting from point A, the time to reach point C is 3.00 s. g = v f - v o /t -9,8ms 2 = v f –(+14.7m/s)/3.0s v f = -14.7m/s

92. d. The distance 0.49 is negative because it is below the origin. Starting from point A, and using the equation 2 gd y = v 2 –v o 2 2(-9.8ms 2 )(-o.49m) = v 2 – (+14.7m/s) 2 = + 15m/s

93. We must choose the negative value because we know that the coin is going down. Therefore, its velocity when it strikes the ground is -15m/s

94. ACCELERATION DUE TO GRAVITY

95. A Free falling body is the most common example of uniformly accelerated motion. The works of Aristotle and Galileo were focused on the description of the motion of a falling body. Based on the results of his study. Aristotle stated that all heavy objects fall faster than light object

96. Galileo suggested that this statement should be tested by an experiment. Galileo was able to arrive at a mathematical description of falling objects. The equations that represents the motion of a freely falling body are written as

97. V = v + gt d = v o t + gt 2 /2 v2 = v o 2 +2gd Where g = acceleration due to gravity v = final velocity vo = initial velocity t = time d = displacement

98. To derived g = acceleration due to gravity d x = vot + 1/2at 2 d y = 1/2gt 2 g = 2dy/t 2

99. EXPERIMENT ACCELERATION DUE TO GRAVITY

100. MATERIALS METAL BALL 0.7M LONG THREAD SOPWATCH METERSTICK

101. PROCEDURE FROM AN ELEVATION OF ABOUT 10M DROP A METAL BALL Record the time it takes the ball to reach the ground Measure the height of fall ( dy ) by dropping one end of the nylon cord until it touches the ground Determine the length of the cord used to measure the height of fall by means of a meterstick . (length of the cord( dy ) Compute for the value of g. using the equation dy = gt 2 /2 Make three trials by repeating steps a to e

102. d y = gt 2 /2 d y = v o t + 1/2gt 2 d y = 1/2gt 2 g = 2dy/t 2

103. DATA AND RESULTS TRIAL d y t g 1 3.7 0.7 2 3. 19 0.60 3 3. 13 0.54 Average

104. DATA AND RESULTS TRIAL d y t g 1 3.7m .47s 2 3.27 .75 3 3.7 .66 Average 3.7 .68

105. COMPUTATION dy = gt 2 g = 2dy/t2 g = 2( 3.7m)/(0.79s)2 = 7.4/.6241 = -11.86m/s2

106. GENERALIZATION IS A SPECIFIC TYPE OF CONCLUSION THAT CAN BE APPLIED MOST IN THE GROUP FROM WHICH THE SAMPLE WAS TAKEN

107. application 1. What two forces act on a body falling through the air

108. Ans :gravity and air resistance