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Video: Analyzing the Kinematics of Decelerating Objects Converging on a Single Point

Ed Burdette

Two trains are moving at 30 m/s 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 m apart. Assuming both trains have the same acceleration, what must this acceleration be if the trains are to stop just short of colliding?

05:53

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 metres 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.