### Video Transcript

Two trains are moving at 30 meters
per second in opposite directions on the same track. The engineers see simultaneously
that they are on a collision course and apply the brakes when they are 1000 meters
apart. Assuming both trains have the same
acceleration, what must this acceleration be if the trains are to stop just short of
colliding?

Let’s start by highlighting some of
the information in the statement. So we’re told that the trains have
a speed, each, of 30 meters per second, and that when they are 1000 meters apart,
the brakes are applied and they start to come to a stop. We’re also told that overall what
we’re looking for is acceleration. We can represent that as just a
single letter 𝑎. That will be the acceleration that
we’re solving for so that these trains stop just short of colliding.

Let’s draw a picture here of what’s
going on. So let’s say we have our first
train moving from left to right, and we’re told that this train has a speed of 30
meters per second, and then we have a second train on the track, same track, moving
in the opposite direction, and that this train has the same speed but the opposite
direction, 30 meters per second. And we’re told further that when
the brakes start to be applied, the trains are a distance of 1000 meters apart from
one another.

With this information, we’re able
to clarify a bit the question that we’re trying to figure or what we’re trying to
solve for. Really what we’d like to figure out
is what is 𝑎, the acceleration, if each train stops in 500 meters?

What we’re looking at here is a
kinematics problem. And because it’s a kinematics
problem, we wanna take a look at the common set of kinematics equations. And here those equations are. You’ll see there’s a set of four
where our 𝑣 sub 𝑓 represents the final velocity, our 𝑣 sub 𝑖 represents initial
velocity, 𝑎 stands in for acceleration, 𝑡 for time, and 𝑑 is our distance
traveled.

Now our next step is to match up
one of these four equations with the information we’re given in this problem. You can see that we aren’t given
any information about time in this problem statement. So when we eliminate those
kinematic equations that involved the variable 𝑡, we’re left with one in terms of
the final velocity, the initial velocity, acceleration, and displacement.

That is the equation we will use to
answer this question of what is 𝑎 if each train stops in five hundred meters. Now let’s take a look at this
equation. So we’ve got a final velocity;
we’ve got an initial velocity; acceleration, that’s what we’re trying to solve for;
and distance which we’re given.

Now if you look at the first term
in this equation, 𝑣 sub 𝑓, we see that that’s actually zero because the trains
final velocity is at rest, zero meters per second. So that simplifies things a
bit. Now what we want to do is
algebraically rearrange this equation so that it will say 𝑎, the variable we’re
interested in solving for, equals something else in terms of what we’re given.

Let’s start by subtracting 𝑣 sub
𝑖 squared from both sides of the equation. You see it cancels out on the right
side of our equation, and if we now divide both sides of the equation by two times
𝑑, or distance traveled, then that also cancels out on the right side of our
equation and we’re left with an equation expressed as 𝑎, our acceleration, equals
negative the initial velocity squared divided by two times our displacement.

Now this is a good place to be
because you can see we’re given what our initial velocity is, 30 meters per second,
and we also know what 𝑑 is. And remember, in this case, we’re
applying this kinematic equation to a single train, which means that the 𝑑 we’ll
use is 500 meters rather than the distance between both trains when they start of
1000 meters. So again, our 𝑑 is 500 meters.

When we plug the given numbers into
this equation, when we plug our given numbers into this equation, we see that we’ve
got negative the square of 30 meters per second divided by two time 500 meters, all
of which equals negative 900 meters squared per second squared divided by 1000
meters.

And when we do this division, we
find that the acceleration of each train is negative zero point nine zero meters per
second squared. That will be the necessary minimum
acceleration of the trains in order to stop before they collide.