1
00:00:03,740 --> 00:00:07,200
If we know the position x of
t of an object as a function

2
00:00:07,200 --> 00:00:09,930
of time, we can
use differentiation

3
00:00:09,930 --> 00:00:12,190
to calculate its velocity
and its acceleration

4
00:00:12,190 --> 00:00:12,960
at later times.

5
00:00:12,960 --> 00:00:15,270
Essentially, by taking the
derivatives of the position,

6
00:00:15,270 --> 00:00:18,910
we know everything there is
to know about the motion.

7
00:00:18,910 --> 00:00:22,060
Sometimes, however, we'll want
to go in the other direction.

8
00:00:22,060 --> 00:00:24,174
We'll have the acceleration
as a function of time

9
00:00:24,174 --> 00:00:26,590
and we'll want to find the
velocity as a function of time,

10
00:00:26,590 --> 00:00:28,680
or the position as
a function of time.

11
00:00:28,680 --> 00:00:31,260
We'll use a technique
called integration,

12
00:00:31,260 --> 00:00:34,200
and let's see how that works.

13
00:00:34,200 --> 00:00:37,500
To begin with, suppose we
have a constant acceleration.

14
00:00:40,530 --> 00:00:48,900
So our acceleration a of
t is some constant a 0.

15
00:00:54,170 --> 00:01:00,300
In that case, we know that
this constant acceleration can

16
00:01:00,300 --> 00:01:03,120
be written as the
change in velocity

17
00:01:03,120 --> 00:01:08,100
delta v over some
time interval delta t,

18
00:01:08,100 --> 00:01:13,940
and therefore that the change in
velocity delta v over some time

19
00:01:13,940 --> 00:01:16,240
interval delta t
can just be written

20
00:01:16,240 --> 00:01:19,670
as a 0 acceleration, the
constant acceleration,

21
00:01:19,670 --> 00:01:22,580
times the elapsed time delta t.

22
00:01:22,580 --> 00:01:25,640
And we can see this
graphically if I

23
00:01:25,640 --> 00:01:28,215
plot acceleration versus time.

24
00:01:30,810 --> 00:01:34,590
Here is my constant
acceleration a 0.

25
00:01:34,590 --> 00:01:38,490
Let's say this is
time 0 and here is

26
00:01:38,490 --> 00:01:44,270
the time delta t, the change
in velocity, a 0 times delta t,

27
00:01:44,270 --> 00:01:54,320
is just the area of this
box, that box being defined

28
00:01:54,320 --> 00:01:57,210
by the time interval
0 to delta t,

29
00:01:57,210 --> 00:02:01,190
and it's the area
under the curve, a 0.

30
00:02:01,190 --> 00:02:04,650
So if we know the
velocity at time 0

31
00:02:04,650 --> 00:02:07,350
and we know the constant
acceleration a 0,

32
00:02:07,350 --> 00:02:11,485
we can calculate the velocity
at any later time in this way,

33
00:02:11,485 --> 00:02:13,860
because the change in velocity
from that initial velocity

34
00:02:13,860 --> 00:02:18,070
is given by a 0 times delta t.

35
00:02:18,070 --> 00:02:20,530
Now let's consider a slightly
more complicated example.

36
00:02:20,530 --> 00:02:23,240
Suppose the acceleration
isn't constant

37
00:02:23,240 --> 00:02:24,821
but is changing linearly.

38
00:02:29,660 --> 00:02:39,760
So for a linearly
changing acceleration,

39
00:02:39,760 --> 00:02:44,570
I can draw this
graphically as well.

40
00:02:44,570 --> 00:02:46,020
Here's acceleration.

41
00:02:46,020 --> 00:02:48,110
This is time equals 0.

42
00:02:48,110 --> 00:02:51,460
Suppose the acceleration
is increasing linearly,

43
00:02:51,460 --> 00:02:57,760
and I'll call this time delta t.

44
00:02:57,760 --> 00:03:00,110
This is a of t.

45
00:03:00,110 --> 00:03:02,860
Note that we can define
an average acceleration

46
00:03:02,860 --> 00:03:05,910
over the interval
from 0 to delta t.

47
00:03:05,910 --> 00:03:11,540
The average acceleration
is the change

48
00:03:11,540 --> 00:03:18,740
in velocity over
the elapsed time.

49
00:03:18,740 --> 00:03:28,720
That average acceleration
will look something like that.

50
00:03:28,720 --> 00:03:32,840
And so then the change
in velocity, delta v,

51
00:03:32,840 --> 00:03:37,560
is equal to the
average acceleration

52
00:03:37,560 --> 00:03:41,760
times the elapsed time.

53
00:03:41,760 --> 00:03:47,770
And this time, notice from
the diagram this product

54
00:03:47,770 --> 00:03:54,579
is actually the area of this
trapezoid that's basically

55
00:03:54,579 --> 00:03:57,030
the area under the a
of t function going

56
00:03:57,030 --> 00:03:58,860
from 0 to delta t.

57
00:03:58,860 --> 00:04:01,610
We've calculated it sort of as
a rectangular area involving

58
00:04:01,610 --> 00:04:03,530
the average acceleration,
but that's also

59
00:04:03,530 --> 00:04:12,550
equal to the area under the time
changing acceleration function.

60
00:04:12,550 --> 00:04:19,459
So this is, again,
the area under a of t

61
00:04:19,459 --> 00:04:21,358
over the time interval delta t.

62
00:04:24,430 --> 00:04:27,830
Now let's consider a more
generally changing acceleration

63
00:04:27,830 --> 00:04:29,150
as a function of time.

64
00:04:29,150 --> 00:04:30,351
So let me plot this.

65
00:04:35,165 --> 00:04:39,880
So suppose we have just some
general function-- I'll draw it

66
00:04:39,880 --> 00:04:44,120
like this-- which
is our acceleration

67
00:04:44,120 --> 00:04:46,200
as a function of time.

68
00:04:46,200 --> 00:04:48,860
The change in
velocity will still

69
00:04:48,860 --> 00:04:52,870
be the area under the curve
a of t over the time interval

70
00:04:52,870 --> 00:04:54,330
we're interested in.

71
00:04:54,330 --> 00:04:56,680
So let's suppose we want
to go from this time

72
00:04:56,680 --> 00:05:01,150
here-- I'll call it t
sub a-- and my ending

73
00:05:01,150 --> 00:05:04,480
time will be t sub b.

74
00:05:04,480 --> 00:05:06,090
So we're interested
in figuring out

75
00:05:06,090 --> 00:05:11,180
what the area under this curve
is over the interval from t sub

76
00:05:11,180 --> 00:05:12,630
a to t sub b.

77
00:05:12,630 --> 00:05:15,010
We can estimate
this by breaking up

78
00:05:15,010 --> 00:05:17,490
the interval into a bunch
of little rectangle.

79
00:05:17,490 --> 00:05:20,510
Suppose I break this up
into n rectangles just

80
00:05:20,510 --> 00:05:21,390
under this curve.

81
00:05:26,140 --> 00:05:31,210
And so this is 1, 2, 3,
4, et cetera, up to n.

82
00:05:31,210 --> 00:05:34,250
And with each one, because
this is a curved graph,

83
00:05:34,250 --> 00:05:38,740
I can't get the rectangle to fit
exactly so a little bit of it

84
00:05:38,740 --> 00:05:42,287
will stick out at
the top of each one.

85
00:05:42,287 --> 00:05:44,370
I'm not going to go through
that for all of these,

86
00:05:44,370 --> 00:05:46,140
but you get the idea.

87
00:05:46,140 --> 00:05:48,290
I'll fit these rectangles
as well as I can.

88
00:05:48,290 --> 00:05:50,790
The narrower I make
the rectangles,

89
00:05:50,790 --> 00:05:53,780
the more easily I can
fit them under the curve.

90
00:05:53,780 --> 00:06:01,680
For each strip, for
each rectangular strip,

91
00:06:01,680 --> 00:06:06,710
let's say from over box
1 from here to here,

92
00:06:06,710 --> 00:06:15,000
the change in velocity is
equal to the average velocity

93
00:06:15,000 --> 00:06:20,010
over that time interval
times the width

94
00:06:20,010 --> 00:06:21,510
of that rectangle, delta t.

95
00:06:21,510 --> 00:06:27,280
So that's the area of the strip.

96
00:06:27,280 --> 00:06:29,230
Essentially what we're
saying is that if we

97
00:06:29,230 --> 00:06:31,690
make the rectangles
narrow enough,

98
00:06:31,690 --> 00:06:36,010
we can treat the acceleration,
the curve of the acceleration,

99
00:06:36,010 --> 00:06:39,200
as roughly constant
over that interval.

100
00:06:39,200 --> 00:06:41,140
And we can make n
as big as we need

101
00:06:41,140 --> 00:06:45,470
to to make those rectangles
very, very narrow.

102
00:06:45,470 --> 00:06:49,540
So in that case, we can estimate
the total change in velocity

103
00:06:49,540 --> 00:06:52,750
from time t sub
a to time t sub b

104
00:06:52,750 --> 00:06:56,830
by adding up the area of all
of these rectangular strips.

105
00:06:56,830 --> 00:06:58,750
So in that case, we write that.

106
00:06:58,750 --> 00:07:01,290
The change in velocity--
so I'll say v sub

107
00:07:01,290 --> 00:07:04,780
b minus v sub a-- that's
the change of velocity going

108
00:07:04,780 --> 00:07:12,330
from time a to time b--
is equal to the sum,

109
00:07:12,330 --> 00:07:19,230
as j goes from 1 to n, of the
area of each of these strips.

110
00:07:19,230 --> 00:07:23,660
And so we want the acceleration
at the i-th rectangle,

111
00:07:23,660 --> 00:07:29,940
so that's a of t sub j times
the width of the strip.

112
00:07:29,940 --> 00:07:31,390
Each strip has the same width.

113
00:07:31,390 --> 00:07:34,640
We'll just call that delta t.

114
00:07:34,640 --> 00:07:42,210
And that's the area of n strips.

115
00:07:42,210 --> 00:07:43,860
Now for a finite--
I really should

116
00:07:43,860 --> 00:07:46,750
write this as
approximately equal to--

117
00:07:46,750 --> 00:07:50,610
because for a finite number
of rectangular strips,

118
00:07:50,610 --> 00:07:53,070
this is just an approximation,
because as I mentioned,

119
00:07:53,070 --> 00:07:56,200
the rectangles don't
exactly fit under the curve.

120
00:07:56,200 --> 00:07:59,190
They can't because the rectangle
doesn't have a curved top,

121
00:07:59,190 --> 00:08:01,610
but the function is curved.

122
00:08:01,610 --> 00:08:04,110
But the narrower I
make the rectangles,

123
00:08:04,110 --> 00:08:07,850
or equivalently,
the larger I make n,

124
00:08:07,850 --> 00:08:09,970
the better the
approximation will be.

125
00:08:09,970 --> 00:08:14,090
So what I want to do is go to
the limit of an infinite number

126
00:08:14,090 --> 00:08:16,720
of rectangles, or
equivalently, the limit

127
00:08:16,720 --> 00:08:19,820
of infinitesimally
narrow strips.

128
00:08:19,820 --> 00:08:22,310
I want delta t to go to 0.

129
00:08:22,310 --> 00:08:24,420
So to make this exact,
what I would write

130
00:08:24,420 --> 00:08:29,330
is that the change in
velocity from time a to time b

131
00:08:29,330 --> 00:08:38,760
is equal to the limit of
the sum from j equals 1 to n

132
00:08:38,760 --> 00:08:45,410
of the acceleration at time t
sub j subject times delta t.

133
00:08:45,410 --> 00:08:50,530
And we want to evaluate that
limit as delta t goes to 0,

134
00:08:50,530 --> 00:08:54,260
or equivalently, as n,
the number of rectangles,

135
00:08:54,260 --> 00:08:57,000
goes to infinity.

136
00:08:57,000 --> 00:09:00,080
Now this is a very
important expression,

137
00:09:00,080 --> 00:09:01,950
and we have a special
way of writing it.

138
00:09:01,950 --> 00:09:05,410
We can also write
it as an integral.

139
00:09:05,410 --> 00:09:09,690
So we write it as the
integral of the function

140
00:09:09,690 --> 00:09:20,500
a of t dt evaluated from
time equals t sub a to time

141
00:09:20,500 --> 00:09:23,350
equals t sub b.

142
00:09:23,350 --> 00:09:35,710
And this is the area under the
a of t curve, the exact area--

143
00:09:35,710 --> 00:09:38,113
not the approximate area, but
the exact area under the a

144
00:09:38,113 --> 00:09:42,060
of t curve over the interval
from t of a to t of b.

145
00:09:42,060 --> 00:09:45,370
This limiting sum
that we've written

146
00:09:45,370 --> 00:09:49,110
this way on the right-hand side
is called the definite integral

147
00:09:49,110 --> 00:09:51,640
of a of t.

148
00:09:51,640 --> 00:09:56,770
And it's related to the
process of integration

149
00:09:56,770 --> 00:09:58,560
that you've learned
about in calculus,

150
00:09:58,560 --> 00:10:01,750
which is the inverse of taking
the derivative, the inverse

151
00:10:01,750 --> 00:10:03,400
of differentiation.

152
00:10:03,400 --> 00:10:06,400
I want to take a moment to
summarize the basic principles

153
00:10:06,400 --> 00:10:09,140
of integration from calculus.

154
00:10:09,140 --> 00:10:11,230
Let's begin by
considering a function

155
00:10:11,230 --> 00:10:14,370
g of x with some derivative.

156
00:10:14,370 --> 00:10:19,889
So consider a function g of x.

157
00:10:19,889 --> 00:10:22,180
And let's assume it's a
well-behaved function, by which

158
00:10:22,180 --> 00:10:24,449
I mean that it's continuous
and differentiable

159
00:10:24,449 --> 00:10:26,240
over the interval that
we're interested in.

160
00:10:26,240 --> 00:10:34,800
So consider g of x
with a derivative.

161
00:10:34,800 --> 00:10:39,820
So dg dx equal to
another function,

162
00:10:39,820 --> 00:10:42,600
which I'll call f of x.

163
00:10:42,600 --> 00:10:46,460
Now note that if I add
a constant to g of x,

164
00:10:46,460 --> 00:10:47,970
I'll still get the
same derivative.

165
00:10:51,040 --> 00:10:53,920
So note that the
derivative with respect

166
00:10:53,920 --> 00:11:04,730
to x of g of x plus
a constant is still

167
00:11:04,730 --> 00:11:06,820
equal to the same
function f of x.

168
00:11:06,820 --> 00:11:09,340
And this is because of the
derivative of a constant

169
00:11:09,340 --> 00:11:12,040
is equal to 0.

170
00:11:12,040 --> 00:11:15,810
Now, suppose I want to
invert this process.

171
00:11:15,810 --> 00:11:18,040
Then I can write that
the antiderivative

172
00:11:18,040 --> 00:11:30,432
of f of x, the
antiderivative of f

173
00:11:30,432 --> 00:11:39,020
of x, which I'll write as
the integral of f of x, dx,

174
00:11:39,020 --> 00:11:48,170
is equal to g of
x plus a constant.

175
00:11:48,170 --> 00:11:51,740
The left-hand side of this we
call an indefinite integral.

176
00:11:58,220 --> 00:12:02,850
And so we see that if
the derivative of g of x

177
00:12:02,850 --> 00:12:07,080
is f of x plus a concert,
the antiderivative of f of x

178
00:12:07,080 --> 00:12:11,460
is g of x plus a constant.

179
00:12:11,460 --> 00:12:14,180
And that can be any
arbitrary constant.

180
00:12:14,180 --> 00:12:17,820
Now in calculus, one
learns how to calculate

181
00:12:17,820 --> 00:12:22,040
the indefinite integral of
various functions, polynomials,

182
00:12:22,040 --> 00:12:24,493
trigonometric functions,
logarithmic functions, et

183
00:12:24,493 --> 00:12:26,840
cetera.

184
00:12:26,840 --> 00:12:30,375
Calculus also shows us how to
compute the definite integral.

185
00:12:34,490 --> 00:12:47,130
So the definite integral,
the integral of f of x, dx,

186
00:12:47,130 --> 00:12:52,330
evaluated from x
equals a to x equals b.

187
00:12:52,330 --> 00:12:54,120
So this is the definite
integral computed

188
00:12:54,120 --> 00:13:04,200
over some interval that is
equal to the antiderivative at x

189
00:13:04,200 --> 00:13:08,190
equals b minus the
antiderivative evaluated

190
00:13:08,190 --> 00:13:10,206
at x equals a.

191
00:13:10,206 --> 00:13:19,750
And this turns out to be
the area under the curve

192
00:13:19,750 --> 00:13:29,910
f of x in the interval between
x equals a and x equals b.

193
00:13:29,910 --> 00:13:32,529
Now notice that there
is no arbitrary constant

194
00:13:32,529 --> 00:13:33,570
in the definite integral.

195
00:13:33,570 --> 00:13:37,180
In the indefinite integral,
we have an arbitrary constant,

196
00:13:37,180 --> 00:13:39,720
but in the definite integral,
that arbitrary constant

197
00:13:39,720 --> 00:13:43,310
is determined by setting
the integration limits.

198
00:13:43,310 --> 00:13:44,720
So there's no
arbitrary constant.

199
00:13:44,720 --> 00:13:47,920
We just have this difference.

200
00:13:47,920 --> 00:13:51,280
And so just to see
this graphically,

201
00:13:51,280 --> 00:14:01,050
if I plot my function
f of x and suppose

202
00:14:01,050 --> 00:14:06,270
this is x equals a and
this is x equals b,

203
00:14:06,270 --> 00:14:08,500
this definite
integral represents

204
00:14:08,500 --> 00:14:15,600
the area under the curve f of
x in the interval from x equals

205
00:14:15,600 --> 00:14:18,080
a to x equals b.

206
00:14:18,080 --> 00:14:21,290
So calculus tells us how
to solve this area problem,

207
00:14:21,290 --> 00:14:23,040
how to compute a
definite integral,

208
00:14:23,040 --> 00:14:24,890
from the antiderivative
that you get

209
00:14:24,890 --> 00:14:27,780
from indefinite integration.

210
00:14:27,780 --> 00:14:30,250
And so this same
technique tells us

211
00:14:30,250 --> 00:14:35,560
how to determine the velocity
from the acceleration,

212
00:14:35,560 --> 00:14:37,880
since we saw that that
was equivalent to an area

213
00:14:37,880 --> 00:14:41,050
under the curve problem.

214
00:14:41,050 --> 00:14:43,110
So to come back to
the motion of objects,

215
00:14:43,110 --> 00:14:47,020
we've shown that the change
in velocity of an object

216
00:14:47,020 --> 00:14:50,980
can be written as the definite
integral of the acceleration.

217
00:14:50,980 --> 00:14:54,070
So just to write that
a little bit more

218
00:14:54,070 --> 00:14:59,270
formally first with
a plot, if this

219
00:14:59,270 --> 00:15:07,400
is my acceleration as
a function of time,

220
00:15:07,400 --> 00:15:12,480
we know that the time
derivative of the velocity

221
00:15:12,480 --> 00:15:17,790
is equal to the acceleration
as a function of time.

222
00:15:17,790 --> 00:15:26,430
I can rewrite that as the
differential dv is equal to a

223
00:15:26,430 --> 00:15:29,200
of t times the differential dt.

224
00:15:32,180 --> 00:15:36,440
And so then I can integrate
both sides of this equation

225
00:15:36,440 --> 00:15:41,920
by writing the
integral over dv is

226
00:15:41,920 --> 00:15:48,370
equal to the integral of the
acceleration of a of t dt.

227
00:15:48,370 --> 00:15:57,080
And I can go from time
equals some initial time t 0

228
00:15:57,080 --> 00:15:59,790
to time equal to
some later time t sub

229
00:15:59,790 --> 00:16:02,979
1 on the right-hand side to
make a definite integral.

230
00:16:02,979 --> 00:16:05,270
And then on the left-hand
side the corresponding limits

231
00:16:05,270 --> 00:16:08,160
are the velocity
at time 0-- I'll

232
00:16:08,160 --> 00:16:14,420
call that v 0-- and the velocity
at time 1, which I'll call v1.

233
00:16:14,420 --> 00:16:16,940
So just to be clear,
I'm assuming here

234
00:16:16,940 --> 00:16:20,920
that v 0 is equal to the
velocity at time t 0,

235
00:16:20,920 --> 00:16:26,490
and v1 is equal to the
velocity at time t sub 1.

236
00:16:26,490 --> 00:16:30,720
So this is the integral of a
constant over an interval of v.

237
00:16:30,720 --> 00:16:35,410
This is an interval of the
acceleration over the time

238
00:16:35,410 --> 00:16:37,140
interval in t.

239
00:16:37,140 --> 00:16:44,260
And so the left-hand side--
this is just v1 minus v 0.

240
00:16:44,260 --> 00:16:47,030
And the right-hand side,
without specifying a of t,

241
00:16:47,030 --> 00:16:49,080
I can't actually
evaluate this integral.

242
00:16:49,080 --> 00:16:51,230
I can't specify what the
antiderivative of a of t

243
00:16:51,230 --> 00:16:54,390
is unless I tell you what
the function a of t is.

244
00:16:54,390 --> 00:16:56,890
So we'll just have to leave
it in terms of an integral.

245
00:16:56,890 --> 00:17:03,820
And so that's just the
integral of a of t dt from t

246
00:17:03,820 --> 00:17:08,160
equals t 0 to t equals t sub 1.

247
00:17:08,160 --> 00:17:10,300
So again, this shows
us that the change

248
00:17:10,300 --> 00:17:14,260
in the velocity from time
t 0 to a later time t1

249
00:17:14,260 --> 00:17:16,719
is equal to the
definite integral of a

250
00:17:16,719 --> 00:17:19,310
of t over that integral.

251
00:17:19,310 --> 00:17:23,260
I can rewrite this in terms
of what the velocity is

252
00:17:23,260 --> 00:17:31,070
at some later time, t1, by
writing the velocity at t1

253
00:17:31,070 --> 00:17:41,640
is equal to v 0 plus the
integral of a of t dt from t

254
00:17:41,640 --> 00:17:45,340
equals t 0 to t equals t1.

255
00:17:45,340 --> 00:17:51,380
Note that T1 is just any
later time after time t

256
00:17:51,380 --> 00:17:53,710
0, where we have the
initial velocity.

257
00:17:53,710 --> 00:17:56,720
So a more convenient way
of writing this function

258
00:17:56,720 --> 00:17:59,130
might be to write the
velocity as a function

259
00:17:59,130 --> 00:18:01,820
of some later time t.

260
00:18:01,820 --> 00:18:03,050
So suppose I were to do that.

261
00:18:03,050 --> 00:18:05,920
I'll just rewrite this
equation replacing t1

262
00:18:05,920 --> 00:18:08,940
with an arbitrary time t.

263
00:18:08,940 --> 00:18:13,360
So I have that v
of t is equal to v

264
00:18:13,360 --> 00:18:28,080
0 plus the integral of a of t dt
from t equals t0 to t equals t.

265
00:18:28,080 --> 00:18:29,950
Now there is
something funny here,

266
00:18:29,950 --> 00:18:34,540
because I have t in the
integration variable here.

267
00:18:34,540 --> 00:18:36,550
But I have t as one
of my limits here as

268
00:18:36,550 --> 00:18:39,980
well, whereas if I look
at this expression here,

269
00:18:39,980 --> 00:18:42,230
there's a difference between
the t in the integration

270
00:18:42,230 --> 00:18:44,150
variable and the limit t1.

271
00:18:44,150 --> 00:18:45,890
They actually represent
different things.

272
00:18:45,890 --> 00:18:49,520
So to keep track of that, the
notation that we generally

273
00:18:49,520 --> 00:18:52,780
use in physics is to call this
integration variable t prime

274
00:18:52,780 --> 00:18:55,980
and so we write this
acceleration of t prime,

275
00:18:55,980 --> 00:19:01,990
dt prime, with the time
t prime going from t 0

276
00:19:01,990 --> 00:19:05,060
to some later time t.

277
00:19:05,060 --> 00:19:07,510
Now one has to be cautious here.

278
00:19:07,510 --> 00:19:11,040
In some fields that
prime on a variable

279
00:19:11,040 --> 00:19:14,610
is used to denote a
derivative, a differentiation.

280
00:19:14,610 --> 00:19:16,480
That's not what it means here.

281
00:19:16,480 --> 00:19:19,360
I'm writing t prime
just to distinguish it

282
00:19:19,360 --> 00:19:22,380
from the specific later
time t that I want

283
00:19:22,380 --> 00:19:24,120
to calculate the velocity at.

284
00:19:24,120 --> 00:19:26,910
And it's worth thinking about
what this expression means.

285
00:19:26,910 --> 00:19:30,340
This equation is identical
to this earlier equation

286
00:19:30,340 --> 00:19:33,077
that we derived except
for a change in notation.

287
00:19:33,077 --> 00:19:35,160
And so let's think about
what that notation means.

288
00:19:35,160 --> 00:19:38,290
t prime here is the
integration variable.

289
00:19:38,290 --> 00:19:39,540
It's a placeholder.

290
00:19:39,540 --> 00:19:41,850
Remember, the integral
here, the definite integral,

291
00:19:41,850 --> 00:19:46,780
represents an infinite sum,
an infinite sum of rectangles

292
00:19:46,780 --> 00:19:54,230
between a time t 0
and a later time t.

293
00:19:54,230 --> 00:19:57,790
And that variable t prime--
actually, what I should do now

294
00:19:57,790 --> 00:20:00,970
is I should call this t prime.

295
00:20:00,970 --> 00:20:06,050
This variable t prime is taking
every value from t 0 to t.

296
00:20:06,050 --> 00:20:09,930
So t prime is representing
the running time variable

297
00:20:09,930 --> 00:20:12,760
for all of our strips
that we're adding up

298
00:20:12,760 --> 00:20:15,160
over this definite integral.

299
00:20:15,160 --> 00:20:16,540
So it's a placeholder variable.

300
00:20:16,540 --> 00:20:18,550
We sometimes call
it a dummy variable.

301
00:20:18,550 --> 00:20:21,780
It's just a placeholder
to represent time,

302
00:20:21,780 --> 00:20:25,520
whereas t, the t in the
limit here without the prime,

303
00:20:25,520 --> 00:20:28,320
represents a specific
choice of later time,

304
00:20:28,320 --> 00:20:32,000
some later time t where we
want to calculate the velocity.

305
00:20:32,000 --> 00:20:36,670
So we know the velocity
at some initial time t 0.

306
00:20:36,670 --> 00:20:40,730
We'd like to know the velocity
at some specific later time t.

307
00:20:40,730 --> 00:20:43,580
And to compute that, we
have to integrate over

308
00:20:43,580 --> 00:20:47,360
all times running from t 0 to t.

309
00:20:47,360 --> 00:20:49,900
And that running
integration variable

310
00:20:49,900 --> 00:20:52,890
we represent as t prime
just to distinguish it

311
00:20:52,890 --> 00:20:55,660
from the specific time
t that we are trying

312
00:20:55,660 --> 00:20:58,120
to compute the velocity for.

313
00:20:58,120 --> 00:20:59,920
So now in just the
same way that we've

314
00:20:59,920 --> 00:21:03,790
obtained the velocity by
integrating the acceleration,

315
00:21:03,790 --> 00:21:04,860
we can integrate again.

316
00:21:04,860 --> 00:21:09,240
We can integrate the velocity
to calculate the position.

317
00:21:09,240 --> 00:21:22,010
So given v of t, we can show
that the position at time t

318
00:21:22,010 --> 00:21:24,810
is equal to the
position at time t

319
00:21:24,810 --> 00:21:30,620
0 plus the integral of
the velocity as a function

320
00:21:30,620 --> 00:21:33,650
of time-- I'll write
this as t prime-- dt

321
00:21:33,650 --> 00:21:40,280
prime going from t prime
equals t 0 to t prime equals t.

322
00:21:40,280 --> 00:21:43,830
So this is exactly analogous
to how we computed velocity

323
00:21:43,830 --> 00:21:44,680
from acceleration.

324
00:21:44,680 --> 00:21:48,620
By integrating a second
time, we can go from velocity

325
00:21:48,620 --> 00:21:50,300
to position.

326
00:21:50,300 --> 00:21:53,410
So once we know the
acceleration a of t,

327
00:21:53,410 --> 00:21:57,300
we can use integration to
compute the velocity v of t

328
00:21:57,300 --> 00:22:01,260
if we know the velocity
at some initial time t 0.

329
00:22:01,260 --> 00:22:05,420
And we can also compute
the position x of t

330
00:22:05,420 --> 00:22:10,300
if we know the initial
position at time t 0, x 0.

331
00:22:10,300 --> 00:22:12,780
So we see that given
the acceleration,

332
00:22:12,780 --> 00:22:15,940
we can recover the
velocity and the position.

333
00:22:15,940 --> 00:22:18,320
And as it happens, from
Newton's Second Law,

334
00:22:18,320 --> 00:22:20,840
if we know the forces
acting on an object,

335
00:22:20,840 --> 00:22:22,970
that gives us the
ability to compute

336
00:22:22,970 --> 00:22:24,332
what the acceleration is.

337
00:22:24,332 --> 00:22:25,790
And then given the
acceleration, we

338
00:22:25,790 --> 00:22:29,690
can use integration to find
the velocity and the position.