1
00:00:00,000 --> 00:00:01,942
- [Instructor] What does distance mean?

2
00:00:01,942 --> 00:00:04,207
In physics, distance is
the total length traveled

3
00:00:04,207 --> 00:00:05,319
for a particular trip.

4
00:00:05,319 --> 00:00:07,729
In other words, if an object
went forward then backward

5
00:00:07,729 --> 00:00:10,286
then forward again, to find
the total distance traveled

6
00:00:10,286 --> 00:00:13,136
you would take this
positive path length add it

7
00:00:13,136 --> 00:00:14,638
to this positive path length

8
00:00:14,638 --> 00:00:17,353
and then add this positive path
length which would give you

9
00:00:17,353 --> 00:00:19,110
the total distance traveled.

10
00:00:19,110 --> 00:00:21,035
That's why a formula for
distance can be thought of

11
00:00:21,035 --> 00:00:21,868
as the summation of all the individual

12
00:00:21,868 --> 00:00:24,509
positive path lengths.

13
00:00:24,509 --> 00:00:27,036
The units for distance are
meters, it is not a vector,

14
00:00:27,036 --> 00:00:31,203
and its always positive or
zero, it can never be negative.

15
00:00:31,203 --> 00:00:33,481
What's an example problem
for distance look like?

16
00:00:33,481 --> 00:00:35,336
Let's say you have this
graph and the question was

17
00:00:35,336 --> 00:00:37,735
what's the distance traveled
for the object in the graph

18
00:00:37,735 --> 00:00:39,745
between zero and six seconds.

19
00:00:39,745 --> 00:00:41,230
So between zero and two seconds,

20
00:00:41,230 --> 00:00:43,415
the object moved forward four meters.

21
00:00:43,415 --> 00:00:45,115
Between two and four seconds,

22
00:00:45,115 --> 00:00:46,822
the object moved backward four meters.

23
00:00:46,822 --> 00:00:48,401
Between four and six seconds,

24
00:00:48,401 --> 00:00:50,666
the object moved backward
another four meters

25
00:00:50,666 --> 00:00:52,922
which means the total
distance would be four meters

26
00:00:52,922 --> 00:00:55,550
plus four meters plus four
meters which would give you

27
00:00:55,550 --> 00:00:58,102
12 meters of distance traveled.

28
00:00:58,102 --> 00:01:00,248
What does displacement mean?

29
00:01:00,248 --> 00:01:02,443
In physics, displacement is
the difference in position.

30
00:01:02,443 --> 00:01:05,792
If an object forward then
backward then forward again,

31
00:01:05,792 --> 00:01:08,355
the displacement could be
represented with an arrow

32
00:01:08,355 --> 00:01:10,450
that points from the
initial position all the way

33
00:01:10,450 --> 00:01:11,913
to the final position.

34
00:01:11,913 --> 00:01:14,030
That's why a formula for
displacement could be thought of

35
00:01:14,030 --> 00:01:16,887
as the final position
minus the initial position.

36
00:01:16,887 --> 00:01:19,697
This delta here represents the change in

37
00:01:19,697 --> 00:01:21,202
which means final minus initial.

38
00:01:21,202 --> 00:01:23,068
The SI units for displacement are meters.

39
00:01:23,068 --> 00:01:25,917
It's a vector which means if
the displacement is leftward

40
00:01:25,917 --> 00:01:28,425
or downward the displacement
can be negative.

41
00:01:28,425 --> 00:01:29,901
In other words, if you end up further left

42
00:01:29,901 --> 00:01:31,068
from where you started,

43
00:01:31,068 --> 00:01:33,503
your horizontal displacement's negative,

44
00:01:33,503 --> 00:01:35,593
and if you end up further
down from where you started

45
00:01:35,593 --> 00:01:37,342
your vertical displacement is negative.

46
00:01:37,342 --> 00:01:39,256
What's an example problem
fro displacement look like?

47
00:01:39,256 --> 00:01:41,007
If you were given this
graph and you were asked

48
00:01:41,007 --> 00:01:42,778
to find the displacement of the object

49
00:01:42,778 --> 00:01:44,377
between zero and six
seconds, you wouldn't care

50
00:01:44,377 --> 00:01:46,273
that the object moved forward and backward

51
00:01:46,273 --> 00:01:47,717
and changed directions.

52
00:01:47,717 --> 00:01:49,525
All that you would focus on
is that the initial position

53
00:01:49,525 --> 00:01:52,478
was zero, the final
position at six seconds

54
00:01:52,478 --> 00:01:53,670
was negative four.

55
00:01:53,670 --> 00:01:56,863
That means the displacement
was negative four meters

56
00:01:56,863 --> 00:02:00,272
since you ended four meters
behind where you started.

57
00:02:00,272 --> 00:02:01,857
What does speed mean?

58
00:02:01,857 --> 00:02:04,139
Speed is a way to measure
how fast something's moving.

59
00:02:04,139 --> 00:02:05,744
It's a scalar.

60
00:02:05,744 --> 00:02:08,020
For our purposes is
really just another way

61
00:02:08,020 --> 00:02:09,784
of saying not a vector.

62
00:02:09,785 --> 00:02:12,439
Speed is defined to be
the distance per time.

63
00:02:12,439 --> 00:02:14,977
You could define average
speed as the distance per time

64
00:02:14,977 --> 00:02:16,503
over a large time interval,

65
00:02:16,503 --> 00:02:18,489
or you could define instantaneous speed

66
00:02:18,489 --> 00:02:19,977
as the rate of distance per time

67
00:02:19,977 --> 00:02:21,769
at a particular moment in time.

68
00:02:21,769 --> 00:02:24,384
In other words, if an object
is taking a long winding path,

69
00:02:24,384 --> 00:02:26,786
you could find the average
speed for the entire trip

70
00:02:26,786 --> 00:02:29,169
by taking the total
distance for the entire trip

71
00:02:29,169 --> 00:02:32,296
and dividing by the time it
took to travel that distance,

72
00:02:32,296 --> 00:02:34,624
or to determine the instantaneous speed,

73
00:02:34,624 --> 00:02:36,730
you could look at an
infinitesimally small distance

74
00:02:36,730 --> 00:02:39,163
and divide by the time it took to travel

75
00:02:39,163 --> 00:02:41,228
that infinitesimally small distance.

76
00:02:41,228 --> 00:02:43,365
The units for speed are meters per second.

77
00:02:43,365 --> 00:02:47,168
It is not a vector, and speed
can only be positive or zero.

78
00:02:47,168 --> 00:02:49,301
Speed can never be negative.

79
00:02:49,301 --> 00:02:51,303
What's an example problem
for speed look like?

80
00:02:51,303 --> 00:02:52,752
If you had this graph and you wanted

81
00:02:52,752 --> 00:02:55,223
to determine the average speed
between zero and six seconds,

82
00:02:55,223 --> 00:02:57,126
you'd have to remember that average speed

83
00:02:57,126 --> 00:02:58,950
is the distance per time.

84
00:02:58,950 --> 00:03:01,206
The distance between zero
and six seconds would be

85
00:03:01,206 --> 00:03:05,049
four meters plus four meters
plus another four meters

86
00:03:05,049 --> 00:03:07,377
would be 12 meters of
distance, and the time it took

87
00:03:07,377 --> 00:03:09,965
to travel that 12 meters
of distance was six seconds

88
00:03:09,965 --> 00:03:13,394
which gives us an average
speed of two meters per second.

89
00:03:13,394 --> 00:03:14,874
What does velocity mean?

90
00:03:14,874 --> 00:03:16,584
Velocity's another way to determine

91
00:03:16,584 --> 00:03:17,791
how fast something's moving,

92
00:03:17,791 --> 00:03:20,152
but this one's a vector
which means velocity

93
00:03:20,152 --> 00:03:22,150
can be positive or negative.

94
00:03:22,150 --> 00:03:25,710
Velocity's defined to be
the displacement per time.

95
00:03:25,710 --> 00:03:27,459
You could define the average velocity

96
00:03:27,459 --> 00:03:29,019
as the displacement per time

97
00:03:29,019 --> 00:03:30,881
over a large time interval,
or you could define

98
00:03:30,881 --> 00:03:33,553
the instantaneous velocity
as the rate of displacement

99
00:03:33,553 --> 00:03:35,871
per time at a particular moment in time.

100
00:03:35,871 --> 00:03:38,652
In other words, if an object
took a long winding path

101
00:03:38,652 --> 00:03:40,652
and you wanted to determine
the average velocity

102
00:03:40,652 --> 00:03:43,133
for the entire trip, you
would take the displacement

103
00:03:43,133 --> 00:03:45,351
for the entire trip and
divide by the time it took

104
00:03:45,351 --> 00:03:46,503
for that displacement,

105
00:03:46,503 --> 00:03:48,360
or if you wanted the
instantaneous velocity

106
00:03:48,360 --> 00:03:50,163
at a particular moment, you would take

107
00:03:50,163 --> 00:03:52,431
an infinitesimal
displacement at that moment

108
00:03:52,431 --> 00:03:55,103
and divide by the time it
took for that displacement.

109
00:03:55,103 --> 00:03:56,978
The units of velocity
are meters per second,

110
00:03:56,978 --> 00:03:58,279
and it is a vector.

111
00:03:58,279 --> 00:03:59,734
That means it can be negative.

112
00:03:59,734 --> 00:04:02,437
If the velocity is directed
leftward or downward,

113
00:04:02,437 --> 00:04:05,725
we typically consider the
velocity to be a negative value.

114
00:04:05,725 --> 00:04:07,778
What's an example problem
for velocity look like?

115
00:04:07,778 --> 00:04:09,042
If you have this graph and you wanted

116
00:04:09,042 --> 00:04:11,240
to determine the average
velocity of the object

117
00:04:11,240 --> 00:04:13,483
between zero and six seconds,
you'd have to remember

118
00:04:13,483 --> 00:04:16,036
that average velocity is
the displacement per time.

119
00:04:16,036 --> 00:04:19,003
Since the object starts
at zero meters and ends

120
00:04:19,004 --> 00:04:22,382
at six seconds at negative
four meters, the displacement

121
00:04:22,382 --> 00:04:24,965
is negative four and the
time it took to travel

122
00:04:24,965 --> 00:04:27,135
that displacement was
six seconds which gives

123
00:04:27,135 --> 00:04:28,113
us an average velocity

124
00:04:28,113 --> 00:04:30,579
of negative two-thirds meters per second

125
00:04:30,579 --> 00:04:32,825
for the time interval
between zero and six.

126
00:04:32,825 --> 00:04:34,545
What does acceleration mean?

127
00:04:34,545 --> 00:04:37,138
Acceleration is the rate
at which velocity changes.

128
00:04:37,138 --> 00:04:39,369
You could change your
velocity by speeding up,

129
00:04:39,369 --> 00:04:42,886
slowing down, or by changing
directions by turning.

130
00:04:42,886 --> 00:04:45,797
It's good to remember that
the acceleration of an object

131
00:04:45,797 --> 00:04:48,130
always points in the same
direction as the net force

132
00:04:48,130 --> 00:04:49,247
on that object.

133
00:04:49,247 --> 00:04:51,197
In other words, if you had a ball that had

134
00:04:51,197 --> 00:04:54,070
an initial velocity v i to
the right and the ball speeds

135
00:04:54,070 --> 00:04:56,592
up to a larger v final to the right,

136
00:04:56,592 --> 00:04:58,650
there must have been acceleration
since the velocity changed

137
00:04:58,650 --> 00:05:01,400
and since acceleration always
points in the direction

138
00:05:01,400 --> 00:05:04,793
of the change in velocity since
this object gained velocity

139
00:05:04,793 --> 00:05:07,310
to the right the acceleration
points to the right

140
00:05:07,310 --> 00:05:09,809
which also means there
must have been a net force

141
00:05:09,809 --> 00:05:12,457
on this ball to the right as well.

142
00:05:12,457 --> 00:05:15,406
The formula for acceleration
is the change in velocity

143
00:05:15,406 --> 00:05:17,972
over the time it took for
that velocity to change.

144
00:05:17,972 --> 00:05:20,211
Since delta means final minus initial,

145
00:05:20,211 --> 00:05:22,002
you could write this as the final velocity

146
00:05:22,002 --> 00:05:25,134
minus the initial velocity
over the time it took

147
00:05:25,134 --> 00:05:26,698
for that velocity to change.

148
00:05:26,698 --> 00:05:29,364
The units of acceleration
are meters per second squared

149
00:05:29,364 --> 00:05:32,236
since acceleration's telling
you how many meters per second

150
00:05:32,236 --> 00:05:35,213
the velocity is changing by per second.

151
00:05:35,213 --> 00:05:38,327
Acceleration is a vector
which means it can be negative

152
00:05:38,327 --> 00:05:41,270
if the acceleration points left or down.

153
00:05:41,270 --> 00:05:42,840
We typically consider that acceleration

154
00:05:42,840 --> 00:05:44,595
to have a negative value.

155
00:05:44,595 --> 00:05:46,984
What's an example problem
for acceleration look like?

156
00:05:46,984 --> 00:05:49,032
Let's determine the two
cases where the acceleration

157
00:05:49,032 --> 00:05:51,436
of this ball points to the right.

158
00:05:51,436 --> 00:05:53,758
If a ball's moving to the
right and it's slowing down,

159
00:05:53,758 --> 00:05:57,464
there's gotta be a net force
to the left that slows it down.

160
00:05:57,464 --> 00:06:00,370
If the net force points to
the left, the acceleration

161
00:06:00,370 --> 00:06:02,575
also points to the left
'cause the net force

162
00:06:02,575 --> 00:06:04,981
and the acceleration always
point in the same direction.

163
00:06:04,981 --> 00:06:07,515
If a ball's moving to the
right and it's speeding up,

164
00:06:07,515 --> 00:06:10,194
there's gotta be a net force
to the right that's speeding

165
00:06:10,194 --> 00:06:12,829
it up that also means
the acceleration points

166
00:06:12,829 --> 00:06:14,124
to the right as well.

167
00:06:14,124 --> 00:06:16,139
If the ball's moving leftward
and it's slowing down,

168
00:06:16,139 --> 00:06:19,542
there's gotta be a net force
to the right slowing it down

169
00:06:19,542 --> 00:06:22,404
which means the acceleration
also points to the right.

170
00:06:22,404 --> 00:06:24,819
If a ball's moving leftward
and it's speeding up,

171
00:06:24,819 --> 00:06:26,600
there's gotta be a net force to the left

172
00:06:26,600 --> 00:06:29,194
speeding the ball up which
means the acceleration

173
00:06:29,194 --> 00:06:30,723
also points to the left.

174
00:06:30,723 --> 00:06:32,712
Since acceleration is always
pointing in the same direction

175
00:06:32,712 --> 00:06:36,433
as the net force, we can see
that B and C are the two cases

176
00:06:36,433 --> 00:06:39,284
where acceleration points to the right.

177
00:06:39,284 --> 00:06:42,018
How do you interpret a
position versus time graph?

178
00:06:42,018 --> 00:06:44,912
Well the value of the graph
is giving you the position X.

179
00:06:44,912 --> 00:06:47,781
In other words, at some time
t, if you measure the value

180
00:06:47,781 --> 00:06:51,545
of the graph on the vertical
axis that value is giving

181
00:06:51,545 --> 00:06:53,811
you the position at that time t.

182
00:06:53,811 --> 00:06:56,161
Why do we care about
position versus time graphs?

183
00:06:56,161 --> 00:06:58,022
One reason we care is that the slope

184
00:06:58,022 --> 00:06:59,912
is gonna equal the velocity of the object.

185
00:06:59,912 --> 00:07:02,943
If you can determine the slope,
you can find the velocity.

186
00:07:02,943 --> 00:07:04,446
How do you find the slope?

187
00:07:04,446 --> 00:07:07,053
You find the region you want
to determine the slope in

188
00:07:07,053 --> 00:07:09,559
and then you take the
rise over the run which is

189
00:07:09,559 --> 00:07:12,667
the change in x over the
change in time for that portion

190
00:07:12,667 --> 00:07:13,592
of the graph.

191
00:07:13,592 --> 00:07:16,654
Since the rise over run here
is displacement per time,

192
00:07:16,654 --> 00:07:18,737
the slope is equal to the velocity.

193
00:07:18,737 --> 00:07:21,871
Also you can tell if there's
acceleration on a position

194
00:07:21,871 --> 00:07:23,446
versus time graph depending on

195
00:07:23,446 --> 00:07:24,967
whether there's curvature or not.

196
00:07:24,967 --> 00:07:26,813
Curvature that looks like a smiley face

197
00:07:26,813 --> 00:07:28,851
represents positive acceleration.

198
00:07:28,851 --> 00:07:30,695
Curvature that looks like a frowny face

199
00:07:30,695 --> 00:07:32,497
represents negative acceleration.

200
00:07:32,497 --> 00:07:35,663
Graphs that have constant slope
would have constant velocity

201
00:07:35,663 --> 00:07:38,018
and zero acceleration.

202
00:07:38,018 --> 00:07:40,212
What would an example
problem with position

203
00:07:40,212 --> 00:07:41,725
versus time graphs look like?

204
00:07:41,725 --> 00:07:43,258
Let's say we have this graph and we wanted

205
00:07:43,258 --> 00:07:45,969
to determine the instantaneous
velocity at seven seconds.

206
00:07:45,969 --> 00:07:47,934
We locate seven seconds.

207
00:07:47,934 --> 00:07:49,417
To find the velocity in this region,

208
00:07:49,417 --> 00:07:51,347
we'll just find the slope in this region.

209
00:07:51,347 --> 00:07:54,151
Since the slope is constant,
we can find the slope

210
00:07:54,151 --> 00:07:56,453
between any two points in this region

211
00:07:56,453 --> 00:07:58,259
choosing six seconds and eight seconds

212
00:07:58,259 --> 00:07:59,917
will be the most convenient.

213
00:07:59,917 --> 00:08:01,376
We'll use rise over run.

214
00:08:01,376 --> 00:08:03,433
The rise in this case would
be the change in position.

215
00:08:03,433 --> 00:08:05,578
The run would be the change in time.

216
00:08:05,578 --> 00:08:08,229
Since between six seconds
and eight seconds,

217
00:08:08,229 --> 00:08:10,725
the graph drops by four
meters the displacement

218
00:08:10,725 --> 00:08:12,532
would be negative four meters.

219
00:08:12,532 --> 00:08:14,903
The time it took for that
to happen i.e. the run

220
00:08:14,903 --> 00:08:16,093
would be two seconds.

221
00:08:16,093 --> 00:08:18,925
Since the slope in this
region is negative two meters

222
00:08:18,925 --> 00:08:22,457
per second that also equals
the velocity in this region.

223
00:08:22,457 --> 00:08:25,827
How do you interpret a
velocity versus time graph?

224
00:08:25,827 --> 00:08:27,929
Well, the first thing you
need to know is that the value

225
00:08:27,929 --> 00:08:30,639
of the graph on the vertical
axis is giving you the value

226
00:08:30,639 --> 00:08:33,832
of the velocity at that
particular moment in time.

227
00:08:33,832 --> 00:08:36,293
Why do we care about
velocity versus time graphs?

228
00:08:36,294 --> 00:08:38,799
One reason we care is that
the value of the slope

229
00:08:38,799 --> 00:08:41,150
at any given moment is gonna
be equal to the acceleration

230
00:08:41,150 --> 00:08:43,496
of the object at that moment.

231
00:08:43,496 --> 00:08:45,090
You find the slope using the definition

232
00:08:45,090 --> 00:08:46,550
which is rise over run.

233
00:08:46,550 --> 00:08:48,983
In this case, the rise is
gonna be the change in v.

234
00:08:48,983 --> 00:08:51,853
The run is gonna be the change
in t and since the change

235
00:08:51,853 --> 00:08:54,139
in velocity over the change
in time is the definition

236
00:08:54,139 --> 00:08:57,185
of acceleration, the
slope on a velocity graph

237
00:08:57,185 --> 00:08:59,257
is equal to the acceleration.

238
00:08:59,257 --> 00:09:03,099
Also the area under any
particular section of the graph

239
00:09:03,099 --> 00:09:06,582
between two times is gonna
equal the displacement

240
00:09:06,582 --> 00:09:08,949
of the object between those two times.

241
00:09:08,949 --> 00:09:11,746
What would an example
problem involving velocity

242
00:09:11,746 --> 00:09:13,245
versus time graphs look like?

243
00:09:13,245 --> 00:09:14,774
Say you had this graph and
you wanted to determine

244
00:09:14,774 --> 00:09:17,124
the acceleration of the
object at two seconds.

245
00:09:17,124 --> 00:09:19,339
The acceleration is gonna
be equal to the slope

246
00:09:19,339 --> 00:09:21,591
at two seconds, so we
need to find the slope

247
00:09:21,591 --> 00:09:22,673
in this region.

248
00:09:22,673 --> 00:09:24,874
Since the slope is constant,
we can find the slope

249
00:09:24,874 --> 00:09:28,126
between any two points
between zero and four seconds.

250
00:09:28,126 --> 00:09:30,350
I'm just gonna pick zero
seconds and four seconds

251
00:09:30,350 --> 00:09:32,728
because they're convenient,
and we'll find the slope

252
00:09:32,728 --> 00:09:34,466
which is always the rise over the run.

253
00:09:34,466 --> 00:09:37,021
The rise in this case is
the change in velocity.

254
00:09:37,021 --> 00:09:38,655
The run is the change in time.

255
00:09:38,655 --> 00:09:40,898
The change in velocity
between zero seconds

256
00:09:40,898 --> 00:09:44,589
and four seconds would be
negative 60 meters per second.

257
00:09:44,589 --> 00:09:47,018
The change in time between
zero and four seconds

258
00:09:47,018 --> 00:09:49,183
would be four seconds.

259
00:09:49,183 --> 00:09:52,181
This gives us slope of negative
15 meters per second squared

260
00:09:52,181 --> 00:09:55,383
which is also the acceleration
in that time period.

261
00:09:55,383 --> 00:09:57,915
If we wanted to determine the
displacement of the object

262
00:09:57,915 --> 00:10:00,389
between four and six seconds,
we can determine the area

263
00:10:00,389 --> 00:10:02,951
under the graph between
four and six seconds,

264
00:10:02,951 --> 00:10:05,724
the area under a rectangle is
gonna be equal to the height

265
00:10:05,724 --> 00:10:07,348
times the width.

266
00:10:07,348 --> 00:10:10,247
The height in this case is
gonna be equal to negative 30

267
00:10:10,247 --> 00:10:13,200
and the width in this case is
gonna be equal to two seconds

268
00:10:13,200 --> 00:10:16,163
which gives us an area
of negative 60 meters

269
00:10:16,163 --> 00:10:18,547
which means the displacement
between four and six seconds

270
00:10:18,547 --> 00:10:20,364
was negative 60 meters.

271
00:10:20,364 --> 00:10:23,614
How do you interpret an
acceleration versus time graph?

272
00:10:23,614 --> 00:10:26,328
Well the value of the vertical axis

273
00:10:26,328 --> 00:10:27,579
is gonna give you the acceleration.

274
00:10:27,579 --> 00:10:29,788
So if you read the graph at
a particular moment in time,

275
00:10:29,788 --> 00:10:32,023
the value of that vertical axis is giving

276
00:10:32,023 --> 00:10:35,225
you the acceleration of the
object at that moment in time.

277
00:10:35,225 --> 00:10:37,994
The slope in this case is
gonna be equal to the jerk

278
00:10:37,994 --> 00:10:40,012
which really isn't asked
about all that much,

279
00:10:40,012 --> 00:10:42,434
but you do need to know
that the change in velocity

280
00:10:42,434 --> 00:10:45,994
is gonna be represented by
the area under this graph.

281
00:10:45,994 --> 00:10:48,431
If you can find the area under
the graph between two times,

282
00:10:48,431 --> 00:10:51,110
that area is gonna equal
the change in velocity

283
00:10:51,110 --> 00:10:54,808
experienced by that object
between those two times.

284
00:10:54,808 --> 00:10:56,577
What would an example problem involving

285
00:10:56,577 --> 00:10:59,176
acceleration versus time graphs look like?

286
00:10:59,176 --> 00:11:01,774
Let's say an object started
from rest at t equals zero,

287
00:11:01,774 --> 00:11:04,391
and you want to determine
what this acceleration graph

288
00:11:04,391 --> 00:11:06,537
what the velocity was at six seconds.

289
00:11:06,537 --> 00:11:09,022
We know that the area is
gonna represent the change

290
00:11:09,022 --> 00:11:09,970
in velocity.

291
00:11:09,970 --> 00:11:12,062
Since I'm going from zero
seconds to six seconds,

292
00:11:12,062 --> 00:11:14,789
I want to determine the
area under the graph

293
00:11:14,789 --> 00:11:17,649
between zero and six
seconds this first area

294
00:11:17,649 --> 00:11:18,998
would be a positive area.

295
00:11:18,998 --> 00:11:21,327
This second area would be a negative area.

296
00:11:21,327 --> 00:11:23,668
This third area would
also be a negative area.

297
00:11:23,668 --> 00:11:26,601
Note that this first
triangular area will cancel

298
00:11:26,601 --> 00:11:28,559
with this second triangular area.

299
00:11:28,559 --> 00:11:31,072
Since one is positive and the
other is equally negative.

300
00:11:31,072 --> 00:11:33,196
The only important area
in this case is gonna be

301
00:11:33,196 --> 00:11:36,216
this rectangle which the
height is negative 10

302
00:11:36,216 --> 00:11:38,589
and the width is two
seconds which gives us

303
00:11:38,589 --> 00:11:41,544
negative 20 meters per
second for the area.

304
00:11:41,544 --> 00:11:44,064
That's also gonna equal
the change in velocity.

305
00:11:44,064 --> 00:11:47,564
Now since the object started
from rest at t equals zero

306
00:11:47,564 --> 00:11:51,034
and it changed its
velocity by negative 20,

307
00:11:51,034 --> 00:11:53,669
the final velocity at six
seconds is gonna equal

308
00:11:53,669 --> 00:11:55,895
negative 20 meters per second.

309
00:11:55,895 --> 00:11:57,673
What are the kinematic formulas?

310
00:11:57,673 --> 00:12:00,084
The kinematic formulas are formulas

311
00:12:00,084 --> 00:12:02,071
that relate the five kinematic variables.

312
00:12:02,071 --> 00:12:04,830
Those variables are
displacement, initial velocity,

313
00:12:04,830 --> 00:12:07,957
final velocity, acceleration, and time,

314
00:12:07,957 --> 00:12:10,345
but these kinematic formulas
only give true relationships

315
00:12:10,345 --> 00:12:13,069
if the acceleration is constant.

316
00:12:13,069 --> 00:12:16,070
These three kinematic formulas
are given on the AP test,

317
00:12:16,070 --> 00:12:18,115
but this last one is not.

318
00:12:18,115 --> 00:12:20,496
One way to remember this formula
is that the left hand side

319
00:12:20,496 --> 00:12:23,528
is v final plus v initial
over two which is the average

320
00:12:23,528 --> 00:12:26,839
of the velocities, and the right hand side

321
00:12:26,839 --> 00:12:28,754
is displacement per time
which is the definition

322
00:12:28,754 --> 00:12:31,480
of average velocity so
it says that the average

323
00:12:31,480 --> 00:12:35,533
of the velocities is equal
to the average velocity.

324
00:12:35,533 --> 00:12:37,698
In other words, during some motion it'll

325
00:12:37,698 --> 00:12:40,001
take a certain time t for an
object to change its velocity

326
00:12:40,001 --> 00:12:43,079
from v initial to v final,
and during that time

327
00:12:43,079 --> 00:12:46,123
the object will have a certain
acceleration and displacement

328
00:12:46,123 --> 00:12:48,643
which will all be
related by these formulas

329
00:12:48,643 --> 00:12:51,110
if that acceleration is constant.

330
00:12:51,110 --> 00:12:52,342
What would an example problem

331
00:12:52,342 --> 00:12:54,057
involving kinematic formulas look like?

332
00:12:54,057 --> 00:12:56,633
Let's say a confused
chipmunk started from rest

333
00:12:56,633 --> 00:13:00,110
and sped up with constant
acceleration for three seconds

334
00:13:00,110 --> 00:13:02,213
traveling nine meters in the process,

335
00:13:02,213 --> 00:13:04,558
we could figure out the
acceleration of the chipmunk

336
00:13:04,558 --> 00:13:06,513
using the second kinematic formula.

337
00:13:06,513 --> 00:13:08,636
The displacement would be nine meters.

338
00:13:08,636 --> 00:13:11,538
The initial velocity would
be zero since the chipmunk

339
00:13:11,538 --> 00:13:12,985
started from rest.

340
00:13:12,985 --> 00:13:14,527
The acceleration is the unknown,

341
00:13:14,527 --> 00:13:17,311
and the time was three
seconds which solving

342
00:13:17,311 --> 00:13:20,589
for the acceleration gives us
two meters per second squared.

343
00:13:20,589 --> 00:13:22,990
We could also solve for the
final speed of the chipmunk

344
00:13:22,990 --> 00:13:24,706
using the first kinematic formula.

345
00:13:24,706 --> 00:13:27,042
The final speed would be our unknown.

346
00:13:27,042 --> 00:13:28,911
The initial speed is still zero.

347
00:13:28,911 --> 00:13:30,803
We know the acceleration is two,

348
00:13:30,803 --> 00:13:33,745
and the time was three seconds
which gives a final velocity

349
00:13:33,745 --> 00:13:35,718
of six meters per second.

350
00:13:35,718 --> 00:13:37,298
The reason we could use these formulas

351
00:13:37,298 --> 00:13:40,495
is because we knew the
acceleration was constant.

352
00:13:40,495 --> 00:13:43,056
A freely falling or flying
object is any object

353
00:13:43,056 --> 00:13:45,160
that has been dropped or
thrown through the air

354
00:13:45,160 --> 00:13:48,257
that only has the force
of gravity acting upon it.

355
00:13:48,257 --> 00:13:50,618
This means we'll typically
ignore air resistance

356
00:13:50,618 --> 00:13:52,489
on these objects unless otherwise stated.

357
00:13:52,489 --> 00:13:54,799
We care about these objects
because the acceleration

358
00:13:54,799 --> 00:13:57,334
in the vertical direction
for all freely falling

359
00:13:57,334 --> 00:13:59,222
and flying objects near the earth

360
00:13:59,222 --> 00:14:01,972
will be negative 9.8
meters per second squared.

361
00:14:01,972 --> 00:14:04,468
Since this acceleration
is gonna be constant

362
00:14:04,468 --> 00:14:07,196
we can use the kinematic
formulas to describe the motion

363
00:14:07,196 --> 00:14:09,835
of freely flying or falling objects.

364
00:14:09,835 --> 00:14:12,870
We just have to plug negative 9.8 in

365
00:14:12,870 --> 00:14:14,655
for the vertical acceleration.

366
00:14:14,655 --> 00:14:15,488
Watch out for code words.

367
00:14:15,488 --> 00:14:19,262
Dropped is code word for
initial velocity is zero.

368
00:14:19,262 --> 00:14:22,361
The final velocity at the
maximum height will be zero.

369
00:14:22,361 --> 00:14:24,713
The acceleration due to gravity
in the vertical direction

370
00:14:24,713 --> 00:14:26,866
for a freely flying object is always

371
00:14:26,866 --> 00:14:29,096
negative 9.8 meters per seconds squared.

372
00:14:29,096 --> 00:14:29,929
Be careful.

373
00:14:29,929 --> 00:14:32,404
If you analyze the motion
of an object flying upward

374
00:14:32,404 --> 00:14:35,283
the velocity gets smaller on the way up

375
00:14:35,283 --> 00:14:37,348
until the velocity is zero at the top.

376
00:14:37,348 --> 00:14:40,528
After which velocity gets
more and more negative

377
00:14:40,528 --> 00:14:42,064
on the way down, however, the acceleration

378
00:14:42,064 --> 00:14:46,231
remains a constant 9.8 the
entire trip on the way up,

379
00:14:47,332 --> 00:14:50,301
on the way down, and even
at the top of the motion

380
00:14:50,301 --> 00:14:52,374
the acceleration and
the vertical direction

381
00:14:52,374 --> 00:14:55,617
is still negative 9.8
meters per second squared.

382
00:14:55,617 --> 00:14:57,104
What would an example problem

383
00:14:57,104 --> 00:14:59,237
involving freely flying objects look like?

384
00:14:59,237 --> 00:15:01,772
Let's say a student drops
a book from a height h,

385
00:15:01,772 --> 00:15:03,972
and we want to know how
long it takes for the book

386
00:15:03,972 --> 00:15:06,561
to hit the ground, but we
need to represent it in terms

387
00:15:06,561 --> 00:15:09,225
of given quantities and
fundamental constants.

388
00:15:09,225 --> 00:15:10,192
That means we're gonna

389
00:15:10,192 --> 00:15:12,018
have to solve this problem symbolically.

390
00:15:12,018 --> 00:15:14,402
We still use a kinematic
formula, but we'll plug

391
00:15:14,402 --> 00:15:17,378
in symbolic values instead of numbers.

392
00:15:17,378 --> 00:15:18,910
Delta y is not gonna be h.

393
00:15:18,910 --> 00:15:21,002
It's gonna be negative h
since the book dropped down

394
00:15:21,002 --> 00:15:23,620
which is a negative vertical displacement.

395
00:15:23,620 --> 00:15:24,914
The book was dropped.

396
00:15:24,914 --> 00:15:27,010
That means the initial velocity was zero.

397
00:15:27,010 --> 00:15:29,595
The acceleration due to
gravity and the y direction

398
00:15:29,595 --> 00:15:31,859
is negative 9.8, but since we're solving

399
00:15:31,859 --> 00:15:34,144
this problem symbolically,
we're not gonna use

400
00:15:34,144 --> 00:15:35,485
an actual number.

401
00:15:35,485 --> 00:15:38,270
We're gonna represent
this as negative little g.

402
00:15:38,270 --> 00:15:41,006
What we mean by little g is the magnitude

403
00:15:41,006 --> 00:15:43,914
of the acceleration due to
gravity which means little g

404
00:15:43,914 --> 00:15:45,247
is positive 9.8.

405
00:15:46,163 --> 00:15:48,727
Since the acceleration due
to gravity is negative 9.8,

406
00:15:48,727 --> 00:15:50,829
we can represent the
acceleration due to gravity

407
00:15:50,829 --> 00:15:52,510
as negative g.

408
00:15:52,510 --> 00:15:55,709
Now we can solve for our time
and we get the square root

409
00:15:55,709 --> 00:15:58,343
of two h over g which
is our symbolic answer

410
00:15:58,343 --> 00:00:00,000
in terms of given quantities
and fundamental constants.