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00:00:03,320 --> 00:00:05,420
You're standing at a
traffic intersection.

2
00:00:05,420 --> 00:00:08,119
And you start to accelerate
when the light turns green.

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00:00:08,119 --> 00:00:12,490
Suppose that your acceleration
as a function of time

4
00:00:12,490 --> 00:00:18,200
is a constant for some time
interval t less than t one.

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And after that, it's zero for
a time after t one less than t

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less than some time t two.

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At the exact same instant
the light turns green,

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a bicyclist is coming
through the intersection.

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And the bicyclist has
some initial speed

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and is braking with an
acceleration of minus

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b two for the entire
time interval t two.

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And at time t two,
the bicyclist comes

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to rest exactly where
you are located.

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And we also know some
initial conditions.

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So our initial conditions
in this problem

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are that you're accelerating
b one at a rate two

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meters per second squared.

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And you do this for time
t one equals one second.

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And the bicyclist comes
into the intersection.

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We'll call that b two naught.

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That's the initial
speed of the bicyclist

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at three meters per second.

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And the question is, what
is the rate of deceleration

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of the bicyclist b two?

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Now this can be quite
a complicated problem.

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So the first thing we want
to do is just make a sketch

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and think about what's involved.

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This problem
involves two objects.

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You and the bicyclist.

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The person-- that's you--
has two stages of motion.

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And the bicyclist only
has one stage of motion.

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So to get started,
it always helps

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to choose a coordinate
system and to make

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some sketches of the problem.

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So let's say we choose a-- it's
all one dimensional motion.

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Two objects.

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One dimensional motion.

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And so we'll pick an origin
at the light at the one

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side of the intersection.

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And we have two objects
which we'll talk about.

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You, x one.

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And the bicycle is x two.

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Actually we don't know yet
who's in front of the other.

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The bicyclist will be
first in front of you.

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So now how do we
sketch the motion

46
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of these two stages of motion?

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So let's make a sketch.

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And let's start with the person.

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Well, the person-- if we
plotted their position

50
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as a function of time-- this
would be position in general.

51
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I'll just draw the
person function.

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They're accelerating
to time t one.

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And then they're moving at a
constant speed at time t two.

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Now the bicyclist is a
little more complicated.

55
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Because initially the bicyclist
has a-- this at time t,

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person one.

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Initially the bicyclist
has a non-zero slope.

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And they're decelerating.

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And they reach you
with a zero slope.

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So this graph,
this is the x two.

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That is the bicyclist.

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And right here we
have the person.

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x one.

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So now to build a strategy,
we can even look at our graph

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and see that from our
initial conditions,

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we have some special conditions
that the-- our strategy will

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be to-- one-- figure
out what this time is.

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And we know that the
bicyclist at time t two

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has come to a stop.

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So that's one condition.

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And we also know
that the bicyclist

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comes to stop exactly at the
same position as the person.

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x one of t two equals
x two of t two.

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So those are two
conditions that we

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can deduce from all of
this given information.

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And now we comply our
kinematic relationships

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for both the bicyclist
and the person

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and try to see if these
conditions will enable

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us to deduce what b two is.

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So let's begin
with the bicyclist.

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So the velocity of the
bicyclist as a function of time

82
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is simply the integration
of that bicyclist b t

83
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prime from zero to t two.

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This is one stage of motion.

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The acceleration is minus b two.

86
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So this is a very
straightforward interval.

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This is just b of two t two.

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b two.

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This is minus the initial
speed equals that.

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b of t two minus V of
the initial is that.

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And because we want
this to be zero,

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we have the condition
that t two equals

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V two naught divided by b two.

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So that's our first
condition for the bicyclist.

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Now we have to separately solve
for the bicyclist's position.

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That's easy.

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x two of t is the integral of V
two t prime dt prime from zero

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to t two.

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And that's just minus one half.

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We want to make sure that
we get the displacement.

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But x two naught is zero.

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So we have x two at t two equals
the integral of the velocity

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function, which is V
two naught minus b two

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of t prime, d t prime
from zero to t two.

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And so we get b two naught
t two minus one half b two t

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two squared.

107
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And when we input this
condition in for t two,

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this becomes very simply V two
naught squared over two b two.

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Substituting t two into
each of these expressions

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gives us that relationship.

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So that's the position of
the cyclist at time t two.

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Now this is a
little bit trickier

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to get the position
of the person.

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So in order to do that,
we first find the velocity

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of the person function.

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It's a two stage motion.

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So for the first stage
of motion, the velocity

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two-- the velocity
of person one--

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minus their initial velocity,
which is zero minus one zero.

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That's zero.

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Equals the integral of
b one dt prime from zero

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to t one, which is
just b one t one.

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And this velocity
remains constant

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throughout the next interval.

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So we can write the velocity
function in the following way.

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V one t equals b one t for zero
less than t less than t one.

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And afterwards, a
constant velocity.

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Now this is the function
that we need to integrate

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to get the displacement.

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So let's get ourselves a little
room here and integrate that.

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And we have x of t
is two integrals.

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First from zero to
t one, the velocity

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function during
that time interval.

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And then for the second
time interval dt two,

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the velocity function is
constant b-- this is b one.

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b one t one, dt prime.

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Notice this is not a variable.

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But it is the time at
the end of the interval.

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And when we make
these two intervals,

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we get one half b
one t one squared.

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00:08:51,560 --> 00:08:58,640
Let's make this the
velocity at time t.

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This first integral
goes from zero to t one.

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And the second
interval, we're going

144
00:09:04,030 --> 00:09:07,620
to make this the
position at time t two.

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00:09:07,620 --> 00:09:15,910
And we get plus V one times t
one times t two minus t one.

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00:09:15,910 --> 00:09:19,880
We have a common term, t
one squared, b, one half b

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00:09:19,880 --> 00:09:20,830
one t one squared.

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00:09:20,830 --> 00:09:23,780
b one minus b one,
t one squared.

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00:09:23,780 --> 00:09:28,820
So this reduces to one
half b one t one squared

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plus b one, t one, t two.

151
00:09:32,870 --> 00:09:34,610
And that's how we
find the position

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of the person for our interval.

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Let's just review
that to make sure.

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Because we had to get the
velocity function first.

155
00:09:43,580 --> 00:09:48,040
And then we integrated the
velocity in each time interval

156
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correctly in order to get
the position function.

157
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Now we can apply our conditions.

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Notice we already
know t two here.

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And we can now apply
the second condition

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which says that the position
of the bicyclist at time t two,

161
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which we found to be d
naught squared over two b two

162
00:10:08,770 --> 00:10:14,090
is equal to the position of
the person at that same time.

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So that's minus one half b one
t one squared plus b one t one.

164
00:10:19,650 --> 00:10:23,530
Now let's make that
substitution for time t two.

165
00:10:23,530 --> 00:10:27,520
So that's V two
naught over b two.

166
00:10:27,520 --> 00:10:33,450
And now our problem is to
solve for this time b two.

167
00:10:33,450 --> 00:10:35,830
And we're given b one.

168
00:10:35,830 --> 00:10:37,530
We're given t one.

169
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We're given V two naught.

170
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And the only variable
here is b two.

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It's a little bit of
algebra to rearrange terms.

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00:10:45,240 --> 00:10:47,910
What I'll do is I'll bring
this term over to here.

173
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So now we'll just do a
little bit of algebra.

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00:10:51,680 --> 00:10:54,410
We have to a b two
we can pull out.

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I have a minus b one
t one b two naught.

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And that's equal to minus
one half b one t one squared.

177
00:11:03,640 --> 00:11:06,350
And now I can solve for b two.

178
00:11:06,350 --> 00:11:10,290
And so I get b two
is equal to Vt naught

179
00:11:10,290 --> 00:11:17,210
squared minus b one t one b two
naught over minus one half b

180
00:11:17,210 --> 00:11:20,590
one t one squared.

181
00:11:20,590 --> 00:11:26,680
Now let's just do a
quick dimensional check.

182
00:11:26,680 --> 00:11:29,750
b times t has the
dimensions of velocity.

183
00:11:29,750 --> 00:11:31,930
So this is velocity
squared, velocity squared.

184
00:11:31,930 --> 00:11:33,570
That's OK upstairs.

185
00:11:33,570 --> 00:11:36,740
b times t squared is
dimensions of position.

186
00:11:36,740 --> 00:11:39,860
So what we have is meters
squared per second squared

187
00:11:39,860 --> 00:11:41,360
divided by meters.

188
00:11:41,360 --> 00:11:43,830
That gives us meters
per second squared.

189
00:11:43,830 --> 00:11:46,800
So we're pretty confident
that we at least didn't

190
00:11:46,800 --> 00:11:48,750
make an algebraic mistake.

191
00:11:48,750 --> 00:11:53,370
And now our last step is to
substitute in the numbers.

192
00:11:53,370 --> 00:11:56,980
And what we get is, if we
put in the three meters

193
00:11:56,980 --> 00:12:03,170
a second squared minus b
one times t one times three,

194
00:12:03,170 --> 00:12:07,710
we get upstairs is minus 3/2.

195
00:12:07,710 --> 00:12:11,710
And downstairs is
two times one second.

196
00:12:11,710 --> 00:12:13,090
Two's cancel.

197
00:12:13,090 --> 00:12:20,380
So we get to 3/2 meters
per second squared

198
00:12:20,380 --> 00:12:23,680
when we put in the numbers.

199
00:12:23,680 --> 00:12:25,930
If we wanted to
check our result,

200
00:12:25,930 --> 00:12:29,456
we can then see
what time we get.

201
00:12:29,456 --> 00:12:41,400
t two is is three meters per
second divided by our b two

202
00:12:41,400 --> 00:12:44,570
which is 3/2 meters
per second squared.

203
00:12:44,570 --> 00:12:47,340
So that's 2 seconds.

204
00:12:47,340 --> 00:12:51,580
And now the last check would
be to see that the position

205
00:12:51,580 --> 00:12:53,530
functions correspond to that.

206
00:12:53,530 --> 00:12:57,440
Let's see if we can just do
that quickly in our heads.

207
00:12:57,440 --> 00:12:59,570
Our position function
for the person

208
00:12:59,570 --> 00:13:04,470
is V naught squared
over two b two.

209
00:13:04,470 --> 00:13:11,110
So that's 9 meters per second.

210
00:13:11,110 --> 00:13:13,660
9 meters squared seconds
squared over two times

211
00:13:13,660 --> 00:13:18,000
3/2 meters second squared.

212
00:13:18,000 --> 00:13:22,590
And that comes
out to x two of t.

213
00:13:22,590 --> 00:13:25,480
This is a check, is 3 meters.

214
00:13:25,480 --> 00:13:30,730
And the x one of t.

215
00:13:30,730 --> 00:13:31,870
We left out the one there.

216
00:13:31,870 --> 00:13:33,450
We should have had it.

217
00:13:33,450 --> 00:13:36,090
It's a little more
complicated to put in here.

218
00:13:36,090 --> 00:13:38,870
But we'll just run the
numbers quickly through.

219
00:13:38,870 --> 00:13:40,730
Two times one seconds minus.

220
00:13:40,730 --> 00:13:42,080
That's a minus one.

221
00:13:42,080 --> 00:13:44,730
Two times one times two.

222
00:13:44,730 --> 00:13:45,860
That's two times two.

223
00:13:45,860 --> 00:13:48,390
So this is also three meters.

224
00:13:48,390 --> 00:13:53,690
And we actually have
the right answer here.