F2 LINEAR MOTION and STATISTICS PPT.pptx

F2 LINEAR MOTION and STATISTICS

1. MATHEMATICS FORM 2 NOTES TOPIC: LINEAR MOTION PREPARED BY: SIR DUKE

2. 17.1: Definitions The distance between two points is the length of the path joining them . The SI unit of distance is the metre . Displacement is distance in a specified direction . Speed is the rate of change of distance with time . The SI unit of speed is m/s , though speed is commonly given in km/h . Velocity is speed in a specified direction or the rate of change of displacement with time. Acceleration on the other hand is the rate of change of velocity with time . The SI unit of acceleration is m/s 2 .

3. Working out Speed Imagine you sailed 45 kilometres between two islands and it took 3 hours. In 3 hours you travel : = 45 km In 1 hour you travel : 45 ÷ 3 = 15 km This means your speed is : 15km/hr Can you guess what the formula is : in symbols

4. Working Example 1 : Liam drove from his house to the Fort William, a distance of 315 km. It took him 3hrs. What was his average speed? Example 2: If a motorist covers a distance of 180 km in 2 hours, Calculate his speed. Working

5. Working out Distance Mixed Problems Example 3: A racing car travelled at 50 km/hr. What is the distance covered in 6 hours 30mins ? Working

6. Working out Time Mixed Problems Example 4 How long did the bus journey take if it travelled a total distance of 100 km at and an average speed of 40 km/hr.

7. Relationship between Time Distance and Speed D S T Simple way to remember the 3 formulae ! D=S T CONCLUSION

8. 17.2: Average Speed Over a time interval, the speed normally varies and what is often used is the average speed. Average speed =  

9. Example 5 A man walks for 40 minutes at 6 km/h. He then travels for two hours in a minibus at 80 km/h. Finally, he travels by bus for one hour at 60 km/h. Find his speed for the whole journey.

10. Example 6 Juma cycled for 2 ½ hours to Mula Trading Centre, 30 km away from his home. He then took his business van and drove at an average speed of 80km/h to a town 180 km away and back to Mula. He finally cycled home at an average speed of 15km/h. Find the average speed for the whole journey. Example 7 A motorist travelled for 2 hours at a speed of 80 km/h before his vehicle broke down. It took him half an hour to repair the vehicle. He then continued with his journey for 1½ hours at a speed of 60 km/h. Determine his average speed.

11. Exercise John travelled from town A to town B, a distance of 30 km. He cycled for 1½ hours at a speed of 16 km/h and then walked the remaining distance at a speed of 3 km/h. Find his average speed for the journey. Anne takes two hours to walk from home to her place of work, a distance of 8 km. On certain day, after walking for 30 minutes, she stopped for ten minutes to talk to a friend. Determine the average speed she should walk to reach her work on time. A motorist drove for 1 hour at 100km/h. She then travelled for 1½ hours at a different speed. If the average speed for the whole journey was 88 km/h, find the average speed for the latter part of the journey. A commuter train moves from station A to station D via stations B and C in that order. The distance from A to C via B is 70 km and that from B to D via C is 88 km. Between the stations A and B, the train travels at an average speed of 48 km/h and takes 15 minutes. Between the stations C and D, the average speed of the train is 45 km/h. Find: the distance from B to C. the time taken between C and D. If the train halts at B for 3 minutes and at C for 4 minutes and its average speed for the whole journey is 50 km/h, find its average speed between B and C.

12. The formula for calculating acceleration is: Acceleration (a) = final velocity ( v f ) – initial velocity (v i ) time (sec) The unit for velocity, in this case, is m/s 2 Velocity and acceleration For motion under constant acceleration; Average velocity = initial velocity + final velocity 2

13. Example 1 A car moving in a given direction under constant acceleration. If its velocity at a certain time is 75  km/h and 1 0 seconds later its 90 km/hr. 

14. Solution Accelaration = change in velocity time taken = ( 90-75 ) km/h 10 s = ( 90- 75 ) x 1000 m/s 2 10 x 60 x 60 5 m/s 2 12 =

15. EXAMPLE 2: A jet starts at rest at the end of a runway and reaches a speed of 80 m/s in 20 s . What is its acceleration?

16. solution Acceleration (a) = final velocity ( v f ) – initial velocity (v i ) time (sec) a = 80 m/s – 0 m/s = 4 m/s 2 20 sec

17. EXAMPLE 3 A skateboarder is moving in a straight line at a speed of 3 m/s and comes to a stop in 2 sec . What is his acceleration?

18. SOLUTION a = 0 m/s - 3 m/s = -1.5 m/s 2 2 m/s

19. Example 4 A car moving with a velocity of 50 km/h then the brakes are applied so that it stops after 20 seconds .

20. NOTE: Negative acceleration is always referred to as deceleration or retardation Acceleration = ( 0- 50 ) x 1000 m/s 2 20 x 60 x 60 = - 25 m/s 2 36 SOLUTION

21. QUESTIONS (1).  The initial velocity of a car is 10 m/s. The velocity of the car after 4 seconds is 30 m/s. Find its acceleration. (2).  A bus accelerates from a velocity of 12 m/s to a velocity of 25 m/s. Find the average velocity during this interval. (3).  A car moves with constant acceleration of 8m/s² for 5 seconds. if the final velocity is 40 m/s, find the initial velocity. (4).  A train driver is moving 40 km/h applies brakes so that there is a constant retardation of 0.5 m/s² . Find the time taken before the train stops.

22. The gradient (slope) of the line on a distance-time graph represents the speed of the object . A steeper gradient means a faster speed . A horizontal line indicates the object is stationary (no change in distance over time). A straight , sloped line indicates constant speed . A curved line indicates changing speed , either accelerating (increasing gradient) or decelerating (decreasing gradient). Distance - Time Graph

23. Distance - Time Graph TIME 9:00am 10:00am 11:00am 11:30am 12:00 noon 1:00pm DISTANCE(KM) 80 160 160 210 310 When distance is plotted against time, a distance-time graph is obtained EXAMPLE 1: The table below shows the distance covered by a motorist from Limuru to Kisumu (a) Draw the distance-time graph (b) Use the graph to answer the following questions: ( i ) How far was the motorist from Limuru at 10.30 a.m ? (ii) What was the average speed during the first part of the journey? (c) What was the average speed for the whole journey?

24. Using Distance Time Graphs The graph ABOVE describes a journey that has several parts to it, each represented by a different straight line. Part A :  09 : 00 − 11 : 00 , the person travelled  30  km away from their starting point and that took them  2  hours. Part B :  11 : 00 − 12 : 00 , we can see that the line is flat, so the distance from their starting point did not change – they were stationary. Part C :  12 : 00 − 12 : 30 , they moved a further  30  km away from their starting point. Part D :  12 : 30 − 14 : 00 , they travelled the full  60  km back to where they began.

25. Interpreting Time – Distance Graphs

26. The graph ABOVE describes a journey that has several parts to it, each represented by a different straight line. Part AB :  10 : 00 − 11 : 00 , the person travelled  6 0 km away from their starting point and that took them  1  hour. Part BC :  11 : 00 − 11 : 3 , we can see that the line is flat, so the distance from their starting point did not change – they were stationary. Part CD :  11 : 3 − 12 : , they moved a further  4 0 km away from their starting point. Part DE :  12 : − 13 : 3 , they travelled the full  10 0 km back to where they began.

27. Distance time graphs –  Key things to remember:  1) The gradient of the line = speed 2) A flat section means no speed ( stopped ) 3) The steeper the graph the greater the speed 4) Negative gradient = returning to start point (coming back)

28. Interpreting Time – Distance Graphs Steady speed (going forward) Stationary (No speed) Steady speed (going backwards) Steady speed (going forward) D( change in distance) T( change in time)

29. Exercise Calculate the speed of the object represented by the green line in the graph, from 0 to 4 s.

30. When velocity is plotted against time , a velocity-time graph is obtained The distance travelled is the area under the graph  The acceleration and deceleration can be found by finding the gradient of the lines .  Velocity- Time Graph

31. Example 1 Find the total distance traveled (2mks) Calculate the acceleration (2mks) Calculate the deceleration (2mks)

32. Example 2 Find the total distance traveled (2mks) Calculate the acceleration (2mks) Calculate the deceleration (2mks)

33. The figure below is a velocity-time graph for a car.  Find the total distance traveled by the car. (2 marks)  b. Calculate the deceleration of the car. (2 marks)  Example 3

34. A particle is projected vertically upwards with a velocity of 30 m/s. If the retardation to motion is 10 m/s², use a graphical method to find the maximum height reached by the particle. A car is travelling at 40 m/s. its brakes are applied and it then decelerates at 8 m/s². Use a velocity-time graph to find the distance it travels before stopping Example 3 Example 4

35. R elative speed is the speed of one object with respect to another. When objects move in the same direction , relative speed is the difference of their speeds .( overtaking bodies ) V relative = V 1 - V 2 When objects move in opposite directions , relative speed is the sum of their speeds .( approaching bodies ) V relative = V 1 + V 2 RELATIVE SPEED

36. Approaching bodies ( V relative = V 1 + V 2 ) Example  A truck left Nyeri at 7.00 am for Nairobi at an average speed of 60 km/h. At 8.00 am a bus left Nairobi for Nyeri at speed of 120 km/h .How far from Nyeri did the vehicles meet if Nyeri is 160 km from Nairobi? 

37. Solution   Distance covered by the lorry in 1 hour = 1 x 60 = 60 km  Distance between the two vehicle at 8.00 am = 160 – 1 00  = 100km  Relative speed = 60 km/h + 120 km/h  Time taken for the vehicle to meet =  ¹⁰⁰ / ₁₈₀   =  ⁵ / ₉  hours  Distance from Nyeri = 60 x  ⁵ / ₉  x 60  = 60 + 33.3  = 93.3 km

38. QUESTION A matatu left town A at 7 a.m. and travelled towards a town B at an average speed of 60 km/h.  A second matatu left town B at 8 a.m. and travelled towards town A at 60 km/h. If the distance between the two towns is 400 km, find;  i . The time at which the two matatus met  ii. The distance of the meeting point from town A  

39. Example  A van left Nairobi for kakamega at an average speed of 80 km/h. After half an hour, a car left Nairobi for Kakamega at a speed of 100 km/h.  a.  Find the relative speed of the two vehicles.  b.  How far from Nairobi did the car over take the van  Overtaking bodies ( V relative = V 1 - V 2 )

40. Solution  Relative speed = difference between the speeds  = 100 – 80 = 20 km/h  Distance covered by the van in 30 minutes  Distance =  ³⁰ / ₆₀  x 80 = 40 km  Time taken for car to overtake matatu =  ⁴⁰ / ₂₀   = 2 hours  Distance from Nairobi = 2 x 100 = 200 km

41. How to solve relative speed; Step 1 . D etermine the direction of motion Determine if the objects are moving in the same direction or opposite directions . Step 2 . Calculate relative speed If moving in the same direction, subtract the smaller speed from the larger speed: V relative = V 1 - V 2 If moving in opposite directions, add their speeds: V relative = V 1 + V 2 CONCLUSION

42. STATISTICS CHAPTER 18

43. Content Definition of statistics Collection and organization of data Frequency distribution tables (for grouped and ungrouped data) Grouping data Mean, mode and median for ungrouped and grouped data Representation of data: line graph, Bar graph, Pie chart, Pictogram, Histogram, Frequency polygon interpretation of data.

44. Definition of statistics This is the branch of mathematics that deals with the collection, organization, representation and interpretation of data .

45. Frequency distribution table A data table that lists a set of scores and their frequency. EXAMPLE 1 The following data represents shoe sizes worn by 20 form two students. Draw a frequency distribution table to represent the data 7,9,6,10,8,8,9,11,8,7,9,6,8,10,9,8,7,7,8,9

46. solution SHOE SIZE TALLY FREQUENCY 6 // 2 7 //// 4 8 //// / 6 9 //// 5 10 // 2 11 / 1 20 7,9,6,10,8,8,9,11,8,7,9,6,8,10,9,8,7,7,8,9 NB/ a) Tally- each stroke is a tally and represents a quantity b) Frequency- number of times an item or value occurs

47. EXAMPLE 2 The following data represents MARKS obtained by 20 form two students in an essay competition. Draw a frequency distribution table to represent the data 9,5,5,4,5,3,5,11,6,3,6,8,9,6,13,8,8,13,5,10

48. solution Marks Tally Frequency(f) 9,5,5,4,5,3,5,11,6,3,6,8,9,6,13,8,8,13,5,10

49. Measures of central Tendency Mean This is usually referred to as arithmetic mean, and is the average value for the data   Where; Known as sigma, means “sum of” represents sum of frequencies represents the sum of the products of f and x  

50. EXAMPLE 3 The following data represents MARKS obtained by 20 form two students in an essay competition. 9,5,5,4,5,3,5,11,6,3,6,8,9,6,13,8,8,13,5,10 Draw a frequency distribution table to represent the data Calculate the mean(average)

51. b) Mode This is the most frequent item or value in a distribution or data. The corresponding frequency is known as modal frequency EXAMPLE 4 From the above Example 3 . State the mode and the modal frequency

52. c) Median When data is arranged in ascending or descending order, the middle item is the median EXAMPLE 5 The following were the marks obtained by 11 students in a mathematics test. Determine the median for the data 40,35,37,50,54,39,60,57,56,51,65

53. NOTE: If there are N items and N is odd , the item occupying position is the median. If there are N items and N is even , the average of items occupying and position is the median.  

54. EXAMPLE 6 The following were the marks obtained by 11 students in a mathematics test. Determine the median for the data 40,35,37,50,54,39,60,57,56,51,65 EXAMPLE 7 The following were the marks obtained by 10 students in a test. Determine the median for the data 7,8,9,14,13,5,10,9,11,6

55. Grouped Data When a large number of items is being considered, it is necessary to group the data. Consider the masses in kilograms of 50 women which were recorded as follows:   63.7 50.0 58.0 73.4 76.3 42.3 58.0 45.0 62.1 78.0 58.0 59.5 45.2 79.6 69.1 42.8 49.8 84.2 54.9 52.6 52.1 63.5 58.0 45.5 58.7 59.5 40.6 61.4 43.5 59.7 72.9 49.2 74.2 60.2 43.9 73.4 53.3 67.3 77.7 63.5 62.3 46.8 66.2 52.1 59.7 60.5 60.3 69.6 80.2 54.4

56. The masses vary from 40.6 kg to 84.2 kg . The difference between the smallest and the biggest values in a set of data is called the range . In this case, the range is 84.2 – 40.6 = 43.6 .

57. The data can be grouped into a convenient number of groups called classes . Usually, the convenient number of classes varies from 6 to 12.

58. How to determine the number of classes Classes= For example, in the above data, a class size of 4 will give = 11 classes. A class size of 5 will give = 9 classes. A class size of 6 will give = 8 classes.  

59. class 40-44 45-49 50-54 55-59 60-64 65-69 70-74 75-79 80-84 frequency  5 5 7 6 13 3 5 3 3 Note: A 40–44 class includes all values equal to or greater than 39.5 but less than 44.5. 40 and 44 are called class limits 39.5 and 44.5 are called class boundaries . The class interval/size (width) is obtained by getting the difference between the class boundaries . In this case, 44.5 – 39.5 = 5 which is the class size/width or class interval .

60. How to get modal class in grouped data The class with the highest frequency is called the modal class . In this case modal class is 60–64 class, with a modal frequency of 13.

61. How to get Median of grouped data For grouped data with N items; The median position is the position if N is odd The median position is the average of the marks in the positions if N is even Median= Lower class boundary + ( x class width )  

62. How to get Mean of grouped data To calculate mean, the midpoint of each class is taken as x an then   Example Class Midpoint(x) Frequency(f) fx 40-44 42.0 5 210 45-49 47.0 5 235 50-54 52.0 7 364 55-59 57.0 6 324 60-64 62.0 13 806 65-69 67.0 3 201 70-74 72.0 5 360 75-79 77.0 3 231 80-84 82.0 3 246 =50 Class Midpoint(x) Frequency(f) fx 40-44 42.0 5 210 45-49 47.0 5 235 50-54 52.0 7 364 55-59 57.0 6 324 60-64 62.0 13 806 65-69 67.0 3 201 70-74 72.0 5 360 75-79 77.0 3 231 80-84 82.0 3 246  

63. Exercise 1. The examination marks in a mathematics test for 72 students were as follows: 36 62 15 28 60 30 25 35 75 14 16 33 58 72 80 92 44 57 55 56 70 34 40 18 15 28 32 60 57 68 83 30 32 40 38 45 48 52 58 62 65 75 84 63 58 47 55 60 38 25 18 29 15 32 35 38 45 43 46 53 58 64 68 54 59 60 61 64 65 70 78 90 Using class interval of 10 and staring from 10–19, make a frequent distribution table. Use the table to: (a) Find the modal class. (b) Estimate: ( i ) the mean. (ii) the median.

64. 2. The heights, to the nearest centimetre , of 50 students in a particular school were as follows: 165 170 182 169 165 180 182 180 184 174 186 175 175 186 174 183 172 167 168 158 188 135 148 159 148 182 163 140 156 158 155 147 143 142 138 160 156 140 148 158 165 180 184 168 Using class interval of 10 and 135 – 144 as the first class, make a frequency distribution table. From the table: state the modal class . estimate the median and the mean .

65. Representation of Data The main purpose of representation of statistical data is to make collected data more easily understood. Methods commonly used in representation are: Bar graphs. Pictograms. Pie charts. Line graphs. Histograms. Frequency polygons .

66. Bar Graphs A bar graph consists of a number of spaced rectangles which generally have their major axes vertical. Bars are of uniform width. The axes must always be labelled and scales indicated. Example Kamau’s performance in five subjects was as follows: Maths 80 English 75 Kiswahili 60 Biology 70 Geography 75 This information can be represented on a bar graph as shown below:

67. Pictograms Data is represented using pictures. Each picture represents a given quantity. Consider the following table which shows the different makes of cars imported over a certain period: Toyota 2 000 Nissan 3 500 Peugeot 1 500 This information can be represented by a pictogram using the following key: Represents 1000 cars Represents 500 cars Toyota Nissan Peugot

68. Question 1 The pictogram shows the number of driving lessons some students had during a month. If Simon had 7 lessons, how many lessons did Samaira have?

69. Solution Simon has three and a half icons. This represents seven lessons. Icon = = 2 Each icon represents two lessons . Samaira has four and a half icons. Samaira = 4.5 × 2 = 9 Samaira had nine lessons.  

70. Question 2 Question 3

71. Pie Chart A pie chart is a circle divided into various sectors. Each sector represents a certain quantity of the item being considered. The size of the sector is proportional to the quantity it represents .

72. Example The total population of animals in a farm given as 1800. Out of these 1200 are chicken, 200 cows, 300 goats and 100 ducks. Represent the information on a pie chart. CHICKEN Angle= x 360 = x 360 =240   DUCKS Angle= x 360 = x 360 =20   GOATS Angle= x 360 = x 360 =60   COWS Angle= x 360 = x 360 =40  

73. 97 98 99 100 101 102 103 6:00 am 8:00 am 9:00 am 10:00 am 11:00 am 7:00 am A hospital nurse recorded a patient’s temperature every hour Example Time Hours Temperature F Temperature ver s us Time Line Graph data is represented using lines.

74. Histograms The frequency in each class is represented by a rectangular bar whose area is proportional to the frequency. Note that the bars are joined together.

75. Histogram when the class interval is the same In this, the frequency in each class is represented by a rectangular bar whose area is proportional to the frequency. When the bars are of same width , the height of the rectangle is proportional to the frequency .

76. Example The following data represents masses to the nearest kilogramme of fish caught by a fisherman in a day. Represent the information on a histogram Mass(kg) 5-9 10-14 15-19 20-24 25-29 30-34 35-39 40-44 No. of fish 6 20 12 10 5 6 2 1

77. Mass(kg) Class boundaries Frequency(f) 5-9 4.5-9.5 6 10-14 9.5-14.5 20 15-19 14.5-19.5 12 20-24 19.5-24.5 10 25-29 24.5-29.5 5 30-34 29.5-34.5 6 35-39 34.5-39.5 2 40-44 39.5-44.5 1

78. b. Histogram when class interval is different Plot frequency density against class boundaries Frequency density( f.d )=   Example Marks 5-9 10-19 20-34 35-39 40-59 frequency 2 10 12 5 4

79. Marks Class boundaries Class width Frequency(f) Frequency density 5-9 4.5-9.5 5 2 0.4 10-19 9.5-19.5 10 10 1 20-34 19.5-34.5 15 12 0.8 35-39 34.5-39.5 5 5 1 40-59 39.5-59.5 20 4 0.2

80. Age 3-5 6-12 13-20 21-34 35-60 frequency 40 78 124 166 252 Exercise 1. Draw a histogram for the data below 2. Draw a histogram for the data below

81. Frequency Polygon This is a graph in which frequency densities are plotted against class mid-points and the points are joined with straight line segments. Normally, the beginning of the graph plotted is extended to start from the previous class (not included in the data provided) with zero frequency. Similarly, the end of the graph plotted is extended to terminate at the next class after the last (not included in the data provided) with zero frequency. Often more than 1 polygon may be drawn on the same axes for purposes of comparing frequency distribution.

82. The masses of 40 students were measured to the nearest kilogram and recorded as shown. MASS(KG) 41-45 46-50 51-55 56-60 61-65 FREQUENCY(f) 5 10 14 8 3 Modify this table to calculate the mean mass. (6 marks) Draw a frequency polygon for the distribution. (4 marks) Example 1

83. Marks 10-19 20-29 30-39 40-49 50-59 60-69 70-79 frequency Maths 2 6 7 13 6 4 2 physics 3 9 10 6 9 2 1 Example 2 1. Draw a frequency polygon for the data below

84. Exercise 1 .The diagram below shows a histogram representing the marks obtained in a certain test:- a. If the frequency of the first class is 20, prepare a frequency distribution table for the data b. State the modal class c. Estimate: i . The mean mark ii. The median mark

85. 2. prepare a frequency distribution table for the data b. State the modal class c. Estimate: i . The mean mark ii. The median mark

86. TOPIC 19: ANGLE PROPERTIES OF A CIRCLE

87. Specific Objectives By the end of the topic the learner should be able to: Identify an arc, chord and segment Relate and compute angle subtended by an arc at the circumference; Relate and compute angle subtended by an arc at the centre and at the circumference State the angle in the semi- circle State the angle properties of a cyclic quadrilateral Find and compute angles of a cyclic quadrilateral.

88. Content a.) Arc, chord and segment. Angle subtended by the same arc at the circumference Relationship between angle subtended at the centre and angle subtended on the circumference by the same arc Angle in a semi- circle Angle properties of a cyclic quadrilateral Finding angles of a cyclic quadrilateral.

89. Definition of a Circle When a  set of all points  that are  at a fixed distance from a fixed point  are joined then the geometrical figure obtained is called circle.

90. Center The fixed point in the circle is called the center. So, the set of points are at a fixed distance from the center of the circle. Radius Radius is the fixed distance between the center and the set of points. It is denoted by  “R” . Diameter The diameter is a line segment, having boundary points of circles as the endpoints and passing through the center. Diameter = Twice the length of the radius or  “D = 2R” Circumference It is the measure of the outside boundary of the circle. So, the  length of the circle or the perimeter of the circle  is called Circumference. Semi-circle A semi-circle is half part of the circle  or, A semi-circle is obtained when a circle is divided into two equal parts.

91. Introduction Arc, Chord and Segment of a circle Arc

92. Arc of a circle The arc of a circle is a portion of the circumference. From any two points that lie on the boundary of the circle, two arcs can be created: A Minor and a Major Arc. Minor arc:  The shorter arc created by two points. Major Arc:  The longer arc created by two points.

93. Chord A chord is a line segment whose endpoints lie on the boundary of the circle.

94. Segment of a circle The SEGMENT of a circle is a portion of a circle created when a chord divides a circle in two regions. Minor segment:  The smaller one created by the chord. Major segment:  The larger one created by the chord.

95. 1. Angles at the Centre and at the Circumference A chord or arc subtends an angle at the centre of a circle. It also subtends an angle at the circumference. However, there is something interesting about these angles. Property:- The angle subtended at the centre is  twice  the angle subtended at the circumference.

96. The animation below shows how this relationship is maintained as the angle at the centre changes.

97. 2 . Angle subtended by the Diameter The diameter subtends an angle of 180 º at the centre of a circle. using the above rule, then the angle subtended at the circumference by the diameter is half i.e 90 º Property:- The angle subtended at the circumference by the diameter is 90 º .

98. The animation below shows how this relationship between the diameter and the angle at the circumference

99. Example 1 Find angle x,y , and z if O is the centre of the circle and < ABC=30 º