1
00:00:01,389 --> 00:00:07,790
Hi guys, how are you today? We are going to continue solving exercises from unit 5.

2
00:00:08,669 --> 00:00:15,710
We are currently on page 120 and I would like to continue with exercise number 13

3
00:00:16,750 --> 00:00:21,710
and the wording says that you leave home and take three minutes to reach a store

4
00:00:22,350 --> 00:00:29,230
15 meters away. If you spend 12 minutes shopping and return the same way in eight minutes,

5
00:00:29,230 --> 00:00:34,429
what was the mean speed of your return and mean of the total journey

6
00:00:34,429 --> 00:00:41,950
okay so we are going to the notebook to start solving this exercise

7
00:00:41,950 --> 00:00:48,350
so let me change it in the tool okay here we are here we are so it's

8
00:00:48,350 --> 00:00:55,950
exercise reset 13 page 120.

9
00:00:55,950 --> 00:01:06,829
So we write the important data we have in this statement, and the first thing is that

10
00:01:06,829 --> 00:01:22,469
in the first motion you take three minutes to cover a space of 500 meters.

11
00:01:22,469 --> 00:01:25,829
That's when you are going in one direction.

12
00:01:25,829 --> 00:01:38,609
Now you stay still without moving for twelve minutes, so in that period of time the distance

13
00:01:38,609 --> 00:01:48,750
that you cover is zero, because you are stopped, and then you return home in eight minutes,

14
00:01:48,750 --> 00:01:56,469
the time for the third part is eight minutes and you cover back the 500

15
00:01:56,469 --> 00:02:05,189
meters right so the first thing we need to know is what is the speed of the

16
00:02:05,189 --> 00:02:13,789
return this is the return okay and to do that so we are going to write the

17
00:02:13,789 --> 00:02:21,270
unknown we have two unknowns in this one is the velocity of the return that is

18
00:02:21,270 --> 00:02:31,129
what what we call v3 and the mean or average speed for all the journey okay

19
00:02:31,129 --> 00:02:38,210
what equation are we going to use okay we will use the equation that defines

20
00:02:38,210 --> 00:02:47,389
the mean speed. Okay, as you know the mean speed is the distance traveled over the

21
00:02:47,389 --> 00:02:52,550
time taken. Okay, the time taken you can write just the t or delta t if it's a

22
00:02:52,550 --> 00:03:01,620
period of time, doesn't matter really. So the solution for the space when you are

23
00:03:01,620 --> 00:03:07,780
returning, so v3 is going to be the distance covered when returning over the

24
00:03:07,780 --> 00:03:16,060
time taken so it's 500 meters over eight minutes but careful before continuing

25
00:03:16,060 --> 00:03:22,180
with these eight minutes you have to convert that into seconds okay to use

26
00:03:22,180 --> 00:03:30,159
international system units okay so we'll do that now that conversion so we write

27
00:03:30,159 --> 00:03:35,939
we divide between minutes and multiply by seconds and we know that one minute

28
00:03:35,939 --> 00:03:50,090
equals 60 60 sorry 60 seconds okay so it's 480 seconds so if you

29
00:03:50,090 --> 00:04:03,370
substitute here you will get a value of 500 over 480 and this is 1.04 meters per

30
00:04:03,370 --> 00:04:15,969
second I'm sorry for that I will modify that okay so 1.04 if we take two decimal places meters per

31
00:04:15,969 --> 00:04:25,509
second so that's the answer to the first question and the second is the mean speed for for the

32
00:04:25,509 --> 00:04:32,829
journey okay so we'll write the mean speed is the total distance covered over

33
00:04:32,829 --> 00:04:40,449
the total time taken and what about the total distance traveled so if going is

34
00:04:40,449 --> 00:04:48,069
500 meters and returning is other 500 meters in total the distance covered is

35
00:04:48,069 --> 00:04:57,910
1000 meters and whatever the time the time is going to be the t1 plus t2 plus

36
00:04:57,910 --> 00:05:09,750
t3 so the three times in total so this is three minutes plus 12 plus 8 okay so

37
00:05:09,750 --> 00:05:18,250
you sum all this up you get the value of 23 minutes and once more we have to

38
00:05:18,250 --> 00:05:27,750
convert this into seconds so we use a conversion factor so minutes seconds 160

39
00:05:27,750 --> 00:05:43,060
we get rid of the minutes and we obtain a value of 1380 seconds and we write

40
00:05:43,060 --> 00:05:52,660
this in the denominator okay so hope you are not getting lost this value is going

41
00:05:52,660 --> 00:06:07,689
to be here and we finally reach to a result of 0.72 meters per second and

42
00:06:07,689 --> 00:06:14,290
this is the total average speed okay so this is the end of exercise number 13

43
00:06:14,290 --> 00:06:26,470
we will proceed now with exercise number 14 next so let me check it okay

44
00:06:26,470 --> 00:06:32,589
exercise number 14 this one you say it says that an airplane takes 2 hours and

45
00:06:32,589 --> 00:06:38,649
18 minutes to complete the journey if its mean speed is 720 kilometers per

46
00:06:38,649 --> 00:06:55,269
how how far did it travel ok so we go to the notebook ok we delete the previous

47
00:06:55,269 --> 00:07:10,860
exercise and we start with exercise we set 14 on page 120 so we have an

48
00:07:10,860 --> 00:07:22,560
airplane the data is that it takes two hours and 18 minutes to cover a distance

49
00:07:22,560 --> 00:07:32,040
and the mean speed of traveling is 720 kilometers per hour okay we have

50
00:07:32,040 --> 00:07:43,019
different choices for this the one choice would be to use only international

51
00:07:43,019 --> 00:07:48,899
system units but this is going to complicate much more the result but you

52
00:07:48,899 --> 00:07:55,560
know it is ok but then you have to to convert everything to kilometers or any

53
00:07:55,560 --> 00:08:03,120
other unit or other thing that we can do is that is to convert this into hours

54
00:08:03,120 --> 00:08:07,199
and then as we have the speed in kilometers per hour we can operate

55
00:08:07,199 --> 00:08:13,759
operate them directly so to do that you know that this is two hours plus 18

56
00:08:13,759 --> 00:08:20,339
minutes agreed so you have two hours and 18 minutes and now we will convert this

57
00:08:20,339 --> 00:08:27,779
18 minutes into hours so this is two hours plus 18 minutes as we want to get

58
00:08:27,779 --> 00:08:32,759
rid of the minutes we will write the minutes in the denominator and the hours

59
00:08:32,759 --> 00:08:40,019
in the numerator as you know one hour is 60 minutes so we write a 1 next to the

60
00:08:40,019 --> 00:08:46,720
hour and a 60 next to the minutes now we can get rid of the minutes and the

61
00:08:46,720 --> 00:08:58,919
number of hours is 18 over 60 which is 0.3 so this is 2 hours plus 0.3 hours so

62
00:08:58,919 --> 00:09:20,159
So, in other words, this is 2.3 hours, right? Did you follow what I did? If you have any doubt, please, you can go to the doubts section and ask or write your question there so that any other student can see your question and benefit from the answer of that question.

63
00:09:20,159 --> 00:09:35,559
Okay, so the unknown is the distance covered and the equation we're going to use is the one that describes the average speed is the distance traveled over the time taken.

64
00:09:35,559 --> 00:09:52,519
As we want to calculate the distance traveled, this time is going to pass to this side multiplying, so you will have the distance traveled equals the velocity or the speed times the time taken.

65
00:09:52,519 --> 00:09:59,480
the velocity is 720 kilometers per hour and as we have the time expressed in

66
00:09:59,480 --> 00:10:05,840
hours now we can cancel hours with hours and we will get the result in kilometers

67
00:10:05,840 --> 00:10:20,679
so 720 times 2.3 and this is 1656 kilometers and that's all okay if you

68
00:10:20,679 --> 00:10:25,960
decide to do it by converting everything into the international system units you

69
00:10:25,960 --> 00:10:30,299
should convert this into seconds and this into meters per second proceed and

70
00:10:30,299 --> 00:10:38,500
then you will get this result in meters which is also right okay next exercise

71
00:10:38,500 --> 00:10:52,320
is exercise number 15 okay we have a graph okay and it says according to the

72
00:10:52,320 --> 00:11:00,000
graph which car moves faster the one represented by the purple line or by the orange line okay so

73
00:11:00,000 --> 00:11:10,299
we are going to solve it in our notebook i'm going to delete this to start from scratch

74
00:11:14,299 --> 00:11:23,580
and we write as usual exercise 15 page 120 okay so we will start by

75
00:11:23,580 --> 00:11:30,720
By drawing the graph, because the graph is the data we have, we don't need to draw it

76
00:11:30,720 --> 00:11:38,019
very accurately because we are not being asked to answer with precise numbers, just to give

77
00:11:38,019 --> 00:11:39,080
a general idea.

78
00:11:39,080 --> 00:11:51,059
So we have a graph where we have a time in hours and the distance covered, or the space,

79
00:11:51,059 --> 00:11:52,059
in kilometers.

80
00:11:52,059 --> 00:11:56,679
have a purple well this is the closer to

81
00:11:56,679 --> 00:11:59,980
purple that I have in these two so a

82
00:11:59,980 --> 00:12:06,519
purple vehicle moving that far and an

83
00:12:06,519 --> 00:12:10,659
orange vehicle moving that far and the

84
00:12:10,659 --> 00:12:14,259
first question is well the first and

85
00:12:14,259 --> 00:12:18,039
only question is which car is moving

86
00:12:18,039 --> 00:12:29,080
faster so we have distance versus time so if you look at at any moment imagine

87
00:12:29,080 --> 00:12:35,940
we check this both at the beginning of the motion were in the origin and after

88
00:12:35,940 --> 00:12:47,200
a certain period of time the orange car has moved this far while the purple car

89
00:12:47,200 --> 00:12:55,840
has moved this far so the purple has covered this distance while the orange

90
00:12:55,840 --> 00:13:03,340
has covered this distance so the purple car is moving faster okay that's one way

91
00:13:03,340 --> 00:13:09,700
of looking at this the other way is as they are moving at a constant speed the

92
00:13:09,700 --> 00:13:16,600
graph of space or distance cover versus time is a straight line and when the

93
00:13:16,600 --> 00:13:26,620
slope okay the slope of of this straight line is larger in the purple one that

94
00:13:26,620 --> 00:13:40,100
means that the purple car is moving faster so the purple moves faster right

95
00:13:40,100 --> 00:13:48,500
now we proceed to exercise number 17 okay so we are going to read the

96
00:13:48,500 --> 00:13:57,080
statement before continuing okay exercise 17 says what is the change in

97
00:13:57,080 --> 00:14:03,440
speed if the acceleration is minus 2.5 meters per second squared for 6 seconds

98
00:14:03,440 --> 00:14:08,899
okay by the way when you have a negative acceleration that means that you're

99
00:14:08,899 --> 00:14:21,080
braking or the speed of the of the mobile is being slower and slower

100
00:14:21,080 --> 00:14:35,850
ok so we have exercise 17 page 120 so we write the data we have is firstly we

101
00:14:35,850 --> 00:14:46,950
We have an acceleration of minus 2.5 meters per second squared, and the time taken for

102
00:14:46,950 --> 00:14:55,190
the braking in this case is 6 seconds, and the unknown that we have to discover or to

103
00:14:55,190 --> 00:15:01,029
calculate is delta V, right?

104
00:15:01,029 --> 00:15:04,110
We know the definition of acceleration.

105
00:15:04,110 --> 00:15:10,509
So that's the equation we are going to use, is the definition of the acceleration.

106
00:15:10,509 --> 00:15:19,509
The acceleration is the change in speed over, or divided by, the time taken in the braking

107
00:15:19,509 --> 00:15:21,870
or acceleration.

108
00:15:21,870 --> 00:15:30,350
So as what is the unknown here is delta v, we need to isolate delta v, or solve for delta

109
00:15:30,350 --> 00:15:31,350
v.

110
00:15:31,350 --> 00:15:39,149
Delta t will pass to the other side of the equality multiplying, and that's how we are

111
00:15:39,149 --> 00:15:40,649
going to solve this.

112
00:15:40,649 --> 00:15:47,269
Okay, regarding the units, we have all the units in the international system, so we won't

113
00:15:47,269 --> 00:15:49,529
have any problem.

114
00:15:49,529 --> 00:15:56,490
And we write that delta v equals the acceleration times delta t as we passed this to this side

115
00:15:56,490 --> 00:15:58,889
multiplying.

116
00:15:58,889 --> 00:16:08,230
Delta V is going to be the acceleration, minus 2.5 meters per second squared, times 6 seconds,

117
00:16:08,230 --> 00:16:16,830
so we cancel the 6 with the square, and then we have 2.5 times 6, this is 15, so minus

118
00:16:16,830 --> 00:16:18,889
15 meters per second.

119
00:16:18,889 --> 00:16:28,769
So in those 6 seconds, the speed decreased 15 meters per second.

120
00:16:28,769 --> 00:16:36,570
So the final speed is 15 meters per second slower than the initial speed.

121
00:16:36,570 --> 00:16:44,889
So this is the exercise 17, and now we are going to cover number 18, that is the last,

122
00:16:44,889 --> 00:16:47,169
but not the least.

123
00:16:47,169 --> 00:16:58,149
we go to the book and then in this exercise 18 you have a graph where we

124
00:16:58,149 --> 00:17:04,289
represent the speed versus the time and we have to describe the motion

125
00:17:04,289 --> 00:17:09,369
represented in the following graph of speed as a function of time

126
00:17:09,369 --> 00:17:14,349
okay by the way this part about simple machines will not be covered this year

127
00:17:14,349 --> 00:17:18,069
because you have covered that part in technology

128
00:17:18,069 --> 00:17:23,829
ok so with this this unit would be finished

129
00:17:23,829 --> 00:17:33,430
ok well so we are back in our notebook and we're going to collect this last

130
00:17:33,430 --> 00:17:48,609
exercise for today. So, we write again, this is exercise 18, page 121, and we

131
00:17:48,609 --> 00:17:54,529
have a graph, so we have a graph, you know that when we have a graph, usually the

132
00:17:54,529 --> 00:18:07,720
graph is the data, and we have a velocity in meters per second represented versus

133
00:18:07,720 --> 00:18:15,859
time in seconds, and we have three different segments. For a segment that is

134
00:18:15,859 --> 00:18:22,720
the velocity is increasing, a second segment where the velocity is steady is

135
00:18:22,720 --> 00:18:29,440
not changing is constant and a third segment where the velocity is

136
00:18:29,440 --> 00:18:38,920
decreasing at a constant rate so with a constant acceleration and well we don't

137
00:18:38,920 --> 00:18:46,119
really need the figures or the numbers for this motion because we just want to

138
00:18:46,119 --> 00:18:53,140
have a rough description of the motion. So in the first part you know that the

139
00:18:53,140 --> 00:19:13,150
velocity is increasing at a constant rate, so the acceleration is constant and

140
00:19:13,150 --> 00:19:21,049
positive because the speed is increasing. In the second segment you can see that

141
00:19:21,049 --> 00:19:31,730
the velocity is constant, it's not changing, it's constant, so the acceleration is zero,

142
00:19:31,730 --> 00:19:41,390
we don't have any acceleration, and eventually the mobile is going to reduce the speed, so

143
00:19:41,390 --> 00:19:53,910
So in the third segment, the velocity is decreasing, I'm going to lower this, is decreasing, so

144
00:19:53,910 --> 00:19:59,150
the acceleration is constant, because it's decreasing at a constant rate, acceleration

145
00:19:59,150 --> 00:20:17,210
constant and is negative sorry for that writing constant and negative okay so

146
00:20:17,210 --> 00:20:23,509
that's all for today don't forget to correct your exercises in your notebook

147
00:20:23,509 --> 00:20:30,509
and upload to the virtual classroom picture of your work so that I can

148
00:20:30,509 --> 00:20:35,849
correct it and check that you are working properly okay so keep on working

149
00:20:35,849 --> 00:20:42,289
hope to see you soon and take care see you bye