WEBVTT

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You're standing at a
traffic intersection.

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And you start to accelerate
when the light turns green.

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Suppose that your acceleration
as a function of time

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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

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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

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

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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.

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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

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is simply the integration
of that bicyclist b t

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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.

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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.

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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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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

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to make this the
position at time t two.

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And we get plus V one times t
one times t two minus t one.

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We have a common term, t
one squared, b, one half b

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one t one squared.

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b one minus b one,
t one squared.

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So this reduces to one
half b one t one squared

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plus b one, t one, t two.

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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.

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And then we integrated the
velocity in each time interval

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correctly in order to get
the position function.

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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,

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

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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.

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Now let's make that
substitution for time t two.

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So that's V two
naught over b two.

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And now our problem is to
solve for this time b two.

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And we're given b one.

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We're given t one.

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We're given V two naught.

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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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What I'll do is I'll bring
this term over to here.

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

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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.

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And now I can solve for b two.

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And so I get b two
is equal to Vt naught

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squared minus b one t one b two
naught over minus one half b

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one t one squared.

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Now let's just do a
quick dimensional check.

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b times t has the
dimensions of velocity.

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So this is velocity
squared, velocity squared.

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That's OK upstairs.

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b times t squared is
dimensions of position.

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So what we have is meters
squared per second squared

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divided by meters.

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That gives us meters
per second squared.

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So we're pretty confident
that we at least didn't

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make an algebraic mistake.

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And now our last step is to
substitute in the numbers.

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And what we get is, if we
put in the three meters

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a second squared minus b
one times t one times three,

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we get upstairs is minus 3/2.

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And downstairs is
two times one second.

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Two's cancel.

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So we get to 3/2 meters
per second squared

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when we put in the numbers.

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If we wanted to
check our result,

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we can then see
what time we get.

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t two is is three meters per
second divided by our b two

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which is 3/2 meters
per second squared.

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So that's 2 seconds.

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And now the last check would
be to see that the position

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functions correspond to that.

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Let's see if we can just do
that quickly in our heads.

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Our position function
for the person

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is V naught squared
over two b two.

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So that's 9 meters per second.

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9 meters squared seconds
squared over two times

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3/2 meters second squared.

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And that comes
out to x two of t.

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This is a check, is 3 meters.

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And the x one of t.

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We left out the one there.

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We should have had it.

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It's a little more
complicated to put in here.

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But we'll just run the
numbers quickly through.

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Two times one seconds minus.

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That's a minus one.

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Two times one times two.

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That's two times two.

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So this is also three meters.

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And we actually have
the right answer here.