An unhinged carriage problem

As part of tutoring high school students in maths and physics, I sometimes run across problems that I find interesting. Here’s a deep dive into one involving an unhinged railway carriage which, at first glance, seems to lack sufficient information.

I find it interesting that some solutions become almost embarrassingly obvious given the right perspective. I also like that there’s usually more than one way to solve a problem. This is one of those situations. Here’s the problem:

A train is moving straight and horizontally at a constant speed. The last carriage is uncoupled from the train and slows with a constant deceleration. The train continues to move at its previous speed. The carriage moves straight and horizontally, eventually coming to a stop after 200 m. Determine the distance the train travelled from the time the carriage was uncoupled to the moment the carriage stopped.

Write the answer in metres (m)

When I first saw this, my instinct was that there wasn’t enough information. The solution requires a very specific answer in metres, yet we’ve not been told the carriage’s deceleration, nor the train’s speed. That struck me as odd.

It turns out that it is possible to calculate a definite value for how far the train travelled before the carriage stopped; it just wasn’t obvious to me from the get-go. This leads to one piece of advice:

Have a go at solving the problem, even if you’re unsure.

There’s another piece of advice that I usually dish out to my students:

Try solving the problem in more than one way. If you get the same answer by a different path, then the answer is more likely to be correct. You might also learn something along the way.

But I’m getting ahead of myself.

Let’s have a go at solving the problem. We’ll first use a geometric argument by drawing the speed versus time graphs for the train and the unhinged carriage. Later, we’ll look at other paths to a solution.

Visualising speed versus time

I’m a big fan of drawing pictures and diagrams to get a feel for a given situation. Interestingly enough, my gut feeling this time wasn’t to draw a train moving along a track (which would have been my usual approach) but to draw graphs of speed versus time.

The question mentions constant speed a couple of times, as well as the uncoupled carriage’s constant deceleration. Thus, my physics-addled mind’s eye straight away saw lines on a speed versus time graph, and I started doodling these on a piece of paper. What follows is a more neatly-drawn¹ version interspersed with explanations.

First, we need some axes and to label them:

The carriage’s initial speed is the same as the train’s speed when it is uncoupled. Let’s call this $v_t$ and mention it on the graph, so that we can refer to it again later.

Since we’re only interested in elapsed time, the time that the carriage is uncoupled from the train happens at zero seconds. The time that the carriage stops, we’ll call $t_1$ . We now label the graph with this information:

We know that the carriage slows with constant deceleration and that deceleration is simply negative acceleration, hence we’ll denote it by $-a$ . Deceleration is a line of negative slope $(-a)$ in the speed versus time graph and connects the time-speed coordinate $(0, v_t)$ with $(t_1, 0)$ . Adding this to the graph, we have:

We also know that the area under a speed versus time graph is the distance travelled. Therefore, we shade this area to highlight it and make a note on the graph that this is equal to $200\ \mathrm{m}$ .

That’s cool. We’ve described, in a graphical form, the uncoupled carriage’s behaviour from when it separated from the train up until it stopped moving.

So what happened to the train over this time? Well, we know that it kept going at the same speed. And we know that constant speed on a speed versus time graph is also a straight line, but with zero slope. In other words, it’s a horizontal line from the time-speed coordinate $(0, v_t)$ to $(t_1, v_t)$ . Let’s add that to our graph:

As with the carriage, the area under its curve is how far the train travelled in this time. We can shade in this area as well:

It’s clear from the graph that the area under the line representing the train’s speed versus time behaviour is enclosed by a rectangle. It’s also clear that the area under the carriage’s curve is enclosed by a triangle, and that the triangle’s hypotenuse bisects the rectangle. In such a case, the triangle’s area is half the rectangle’s area. Since we know that the triangle’s area is equal to $200\ \mathrm{m}$ , we can say that the rectangle’s area must be equal to double that, i.e. $400\ \mathrm{m}$ . Thus, the train travelled $400\ \mathrm{m}$ in the time it took the uncoupled carriage to come to a halt.

There we have it! That’s the answer!

Now we can see that it wasn’t important how fast the carriage decelerated, nor was it important how fast the train was travelling. Thus, my initial feeling about missing information was unfounded. Hence, we see why it was a good idea to attempt to solve the problem to see if we could get anywhere. Also, the answer sort of “pops out” at us as soon as we draw the speed versus time graphs: we only needed a bit of geometry and to know the physical meaning of the area under the curve to find the answer. It’s almost blindingly obvious in the end.

We now put our fingers in our ears and go “la! la! la!” loudly, to try and ignore the fact that we know the answer. Then, we’ll solve the problem algebraically without using any pictures or graphs.

Why do this “the hard way”? Why not? Come on, let’s dig in and see what happens!

An algebraic approach

As with the graphical approach above, we collect what we know from the problem description and see if we can make some headway by combining things logically.

We know that the train travels at a constant speed the entire time, so let’s call that $v_t$ . We’re told that the last carriage is unhinged (i.e. uncoupled) from the train at some time; thereafter, it decelerates at a constant rate. Since deceleration is negative acceleration, let’s call this quantity $-a$ . The carriage starts with the same speed as the train, finally coming to a stop at some later time. Let’s set the initial time to zero and call the time the carriage stops $t_1$ . With this information, we can write an equation to describe the carriage’s speed between these two times. Let’s call the carriage’s speed $v_c$ , which we can write as:

where $t$ is the elapsed time.

Note that this equation has the same form as the equation for a straight line in the plane:

where $y = v_c$ , $m = -a$ , $x = t$ and $c = v_t$ .

We can see that (1) behaves the way we want by considering the start and end points.

At the uncoupling event, $t = 0$ , hence when we substitute this into (1), we find that the speed of the carriage when it gets unhinged is:

which is the $y$ intercept. That makes sense, because we know that the carriage’s initial speed is the same as the train’s speed. Cool, that means we’re probably on the right track.

The $-a t$ term in (1) linearly decreases the value of $v_c$ , tracing out a line which then intersects the $x$ -axis at $t_1$ . In other words, the speed is equal to zero at $t_1$ . This is the behaviour we want at the other end of the time interval. Good, we’re still on the right track.

We know that the carriage stops at $t_1$ , thus we have the relationship:

That might not look that useful, but we can rearrange it to give an expression for the time when the carriage stops, and that could come in handy later:

What else do we know? Well, we know that the train travels at constant speed between $t = 0$ and $t = t_1$ . That means we know how far the train went in that time: this is its speed multiplied by the elapsed time. In other words, the distance the train travelled (which we’ll call $d_t$ ) is:

Since we have an expression for $t_1$ (Equation (4)), we can rewrite the expression for $d_t$ to see if it might be useful:

Hrm, that doesn’t look like it’s going to be of much help. Oh well. Let’s add it to the growing pile of things that we know anyway and see what else we know and/or can deduce.

What about the distance the last carriage travels? We know that its value is $200\ \mathrm{m}$ . Can we write an expression for it in terms of the other things we know? Yes, we can! We know that the carriage’s speed is described by:

We also know that distance is the integral of the speed between two points in time.² Thus, to find the distance travelled by the carriage, we integrate Equation (1) with respect to time between the start and end points, $t = 0$ and $t = t_1$ . To be consistent with our other notation, let’s call the distance travelled by the carriage $d_c$ . Putting this all together, we have:

Squinting at this result, one gets the feeling that it’d be nice to remove at least one of the variables. Since we already have an expression for $t_1$ (Equation (4)), we can try substituting that into the equation for $d_c$ to see what happens:

Hrm, that $\frac{v_t^2}{a}$ bit looks familiar. Hang on! That’s the same as the expression for the distance the train travelled from earlier (Equation (7)). In other words,

This means that the carriage travelled half the distance that the train travelled. Which is the same as saying that the train travelled twice the distance that the carriage travelled. Wow, that turned out to be a very simple relationship in the end!

Now, we know that $d_c = 200\ \mathrm{m}$ , as that was given to us in the question. Thus, the distance the train travelled in the time it took for the carriage to come to a stop is $d_t = 400\ \mathrm{m}$ .

And that’s it! We solved the problem. Yay! And we got the same result as the graphical approach outlined earlier. Cool! 🎉 I love it when a calculation comes together. 😉

There’s something else one can learn here:

What looks like a dead end in a calculation might come in useful anyway.

After all, calculating the distance travelled by the train in terms of its velocity and the carriage’s deceleration looked like it wasn’t very helpful. But in the end, it gave us a useful, simplifying hint which led to the answer.

An average solution

There’s another way of looking at this problem. We can consider average speeds.

The fundamental insight with this approach is that, for a situation involving constant acceleration, the distance travelled is proportional to the average speed.

Let’s first consider the case for the train. Its speed at both the beginning ( $t = 0$ ) and at the end ( $t = t_1$ ) is equal to $v_t$ . We know that the distance it travels is its speed multiplied by the elapsed time:

Because the train doesn’t change speed, that means the train’s average speed, $\bar{v}_t$ , is the same as the train’s speed. I.e.

where $v_{t,0}$ and $v_{t,1}$ are the speeds of the train at the beginning and end of the time interval, respectively.

We can now write the distance the train travelled in terms of its average speed:

This can be formulated more loosely as:

Because we’re considering the same time interval for both the distance travelled by the train and by the unhinged carriage, the elapsed time (which is equal to $t_1$ ) ends up being just a scaling constant. Hence, the distance travelled is proportional to the average speed.

Of course, the carriage does change speed, and over the same time interval. We also have a constant acceleration³ for the unhinged carriage, so we have the same kind of proportionality relationship between the distance and average speed, i.e.:

The average speed for the carriage is:

where $v_{c,0}$ and $v_{c,1}$ are the speeds of the carriage at the beginning and end of the time interval, respectively.

Because the carriage’s speed at the beginning is equal to the train’s speed, and the carriage’s speed at the end is equal to zero, we have $v_{c,0} = v_t$ and $v_{c,1} = 0$ . Thus, the carriage’s average speed ends up being:

I’m sure you can see where this is going. But let’s continue anyway to nail everything down more firmly.

Since the distances are proportional to the average speeds, the ratios of the distances will be equal to the ratios of the respective average speeds. In other words, we can write:

Since we know what $d_c$ , $\bar{v}_t$ and $\bar{v}_c$ are, we can rearrange this to give us our only unknown, $d_t$ :

And since we know that $d_c = 200\ \mathrm{m}$ , we have the distance the train travels to be

which is our result from before. Yay!

Wrapping up

So there we have it. Although the solution wasn’t very mind-blowing, the journey turned out to be much more instructive than the destination. And that’s much more valuable from a problem-solving perspective.

1. Why, yes, I did create these graphs with TikZ. Why do you ask? ↩

2. Yes, I’m slyly re-using the idea of the distance being the area under the curve here. ↩

3. Well, it’s a deceleration, but that’s just a negative acceleration. I might have mentioned this before. 😉 ↩