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PROFESSOR: We all
regularly drive with a car.

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We want to get to from
point A to point B. Usually,

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we like to drive with
a constant speed.

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But what happens if
there's a traffic light?

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Well, we have to
stop eventually,

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which means we have to brake.

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And that makes the
car decelerate.

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So there are different parts
of motion when we ride a car.

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It can be constant.

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It can be accelerating.

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It can be decelerating.

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So let's consider this
in a little graphic.

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We have a little car here.

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And it drives with
constant speed

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until someone slams
into the brakes,

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and the car will
eventually stop.

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Let's add a timeline to this.

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Here we're going
to put t equals 0.

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The point in time
where the car stops

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driving with a
constant speed is t1,

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and the point where the car has
come to a complete stop is t2.

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And the problem now asks for
what is the position function?

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What is the distance at
any kind of time here

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as the car goes through these
two different phases of motion?

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So ultimately we want
to get to x of t.

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But we can't actually
just write this up.

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We need to first look
at other information

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that we have about
this kind of set up.

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And what we have is information
on the acceleration.

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And we have to consider
these two phases separately.

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So we have here phase
one and phase two.

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And phase one is from 0 to t1.

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And phase two is from t1 to t2.

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And we know when a car is
going with a constant velocity,

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it's acceleration
is going to be 0.

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So that keeps life
easy for phase one.

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And then, when it breaks,
well, we know it decelerates.

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So we're definitely going
to have a minus sign.

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And it breaks in
between t and t1.

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And we need to multiply this
factor here with a constant.

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So this is our
acceleration information

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that the problem has given us.

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And we know, of course, that if
we integrate the acceleration,

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we're going to get to
the velocity function.

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And if we integrate
that one again,

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we're going to get to
our position function.

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So let's do that
first for phase one.

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And what is also
important here is

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that we need to
carefully consider

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the initial conditions.

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Again, what
additional information

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is this problem giving us?

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So initial conditions-- they
always make life a lot easier.

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So it's important to
carefully look for them.

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And what that means
is we want to know

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what the distance
is at t equal 0,

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and what the velocity
at time 0 is.

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And we know the
distance here is 0.

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And we know that the problem
starts when the car is already

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driving with a constant speed.

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So that's going to be v not.

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So let's then make
use of this, and start

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integrating our acceleration.

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We want to get to
the velocities.

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So v of t minus of t equals 0
is going to be that integral.

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And here already we know that we
have v of t equals 0 is v not.

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So this already
looks pretty nice.

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And now we have our
integral here of a.

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And of course, now t is
an integration variable.

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So we have to give this a prime.

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And we're integrating
from t prime equals 0

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to t prime equals t.

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And when we plug
this in, well, a

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is 0 in phase one,
which means actually,

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our integral is going to be 0.

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That's handy.

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And from that, we will find
that v of t equals simply v not.

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All good.

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Now we need to integrate this
to get to our position function.

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Same thing again-- x of t
minus x of t equals zero.

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Let's quickly check here what
we know about x of t is 0.

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Well, it's 0 from the
initial conditions.

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And now we have to integrate
our velocity function here,

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v not dt, again, prime.

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t prime equals 0.

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t prime equals t.

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These are our boundaries.

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And this is simply going to
be v not t prime evaluated

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from 0 to t.

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And we can add that--

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we can add that in there.

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And it's just going to be vt.

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So x of t--

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I can just write it here
again, x of t equals v not t.

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OK.

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Good.

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So half of the question is--

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half of the problem is solved.

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Let's look at the same
things for phase two.

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So let's consider the initial
conditions for phase two now.

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And again, now we want to--

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in an analogous way--
look at the distance

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of what we know about
the distance at t1

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and the velocity at t1.

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And about the position,
we know from over there

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that this is v not t.

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So our previous result
now actually becomes

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the initial condition
for the next step, right?

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Because whatever
happens here sets

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whatever happens afterwards.

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So that's nice.

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And we know that, up
to this moment here,

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this car goes with
the velocity of v not.