# High school physics problem - having trouble understanding

This is a fairly straightforward problem which doesn't require the usage of more than one or two formula but I find it hard to grasp the concept behind this.

Let's say we have two trains, one which moves at the speed of $$45 \frac{km}{h}$$ and the other at the speed of $$60\frac{km}{h}$$. Now, let the first train start moving, and let the second one start moving an hour after the first one. The question is after how many hours will the second train catch up to the first one.

I have always had trouble visualizing these kind of problems. I know that the second train starts with a delay of $$1$$ hour and that during that time the first train passes $$45$$km. But how do I calculate this?

I know that $$v_2 - v_1 = 15\frac{km}{h}$$ which is the relative speed of the second train with respect to the first one. This probably means that in such a frame of reference, the $$v_1$$ is zero so we can imagine it as being static, under the condition that the new $$v_2=15 \frac{km}{h}$$.

But how do I calculate this? $$t=\frac{s}{v}$$, thus I need a length in order to calculate this. I can't simply plug in the $$45$$ km from above because that would be the time in which the second train got to the $$45$$km mark, but the first train would have moved away from that point. Could anyone explain?

• Note: If you plot the positions of the two trains as functions of time, you'll get two straight lines. The point where those lines intersect tells you when and where the second train catches up to the first. In other words, the problem from analytic geometry, of finding the intersection of two lines, is isomorphic to the problem of computing when the trains meet. Nov 17, 2021 at 15:04
• Are the two trains starting at the same point? That seems to be implied, but I wanted to confirm. Nov 17, 2021 at 21:57
• Don't try to visualize it without drawing a picture! Draw pictures at critical time values. That will help. Nov 17, 2021 at 22:39

After the 1st hour, the first train would have moved 45 km, the second train will just start moving. Therefore, the distance between the two trains is 45 km.

In the next hour, the second train moves 60km, the first train moves only 45 km, so the second train would have caught up by 15 km. The distance between the two trains is now 30km.

Repeat this for another 2 hours and the second train will have caught up with the first train.

The best way to see this will be to draw out the distance travelled by each train per hour, or the sketch the displacement-time graph for each train. The idea of 'relative velocity' is a more complicated way to reason the above situation, although it can be very useful in certain problems.

You can plug in $$45$$ km/h when using the relative speed. Relative speed is the rate the relative distance changes. Let's say the train that remains stationary in the relative frame is train $$0$$, and the other we call train $$1$$. Now go back to the original frame, and set the origin of time $$t=0$$ to be the time that train $$0$$ starts and train $$1$$ is $$45$$ km away. We have $$x_0 = v_0t$$ and $$x_1 = v_1t +45$$ Now subtract the first from the second $$x_1-x_2 = (v_1 - v_2)t + 45$$ The expression on the left is the relative distance. When they catch up, $$x_1=x_2$$.

Your choice to consider the problem in a different frame is insightful. I think most beginners would not think of doing that.

There are three useful frames of reference we could use to study this problem: one that holds the train station static, one that holds the first train (Train A) static, and one that holds the second train (Train B) static.

• From the frame of reference where the station is static/fixed:

• For the first hour, Train A moves at 45 km/h, while Train B moves at 0 km/h.
• After the first hour, Train A continues to move at 45 km/h and Train B moves at 60 km/h.

This is the frame of reference implied by the phrasing of the question, but it is no more or less valid than the other frames of reference.

• From the frame of reference where Train A is static/fixed:

• For the first hour, both the station and Train B move at -45 km/h (that is, they move "backwards").
• After the first hour, the station continues to move at -45 km/h, and Train B moves at 15 km/h (-45 + 60 = 15).
• From the frame of reference where Train B is static/fixed:

• For the first hour, the station moves at 0 km/h, while Train A moves at 45 km/h.
• After the first hour, the station moves at -60 km/h (0 - 60 = -60), while Train A moves at -15 km/h (45 - 60 = -15).

In all three frames of reference, it is easy for us to see that the trains are 45 km apart at the end of the first hour (in either the positive or negative direction depending on the frame of reference). ±45 km/h * 1h = ±45 km.

The movement of the trains after the first hour is best understood as a new sub-problem: "If two trains start 45 km apart, and move at these speeds, how long will it be until they meet?" The solution is not intuitive from the station-based frame of reference, but becomes much more obvious in either of the train-based frames of reference.

Using the t = s/v formula that you mentioned: s/v is (45 km)/(15 km/h) for Train A's frame, or (-45 km)/(-15 km/h) for Train B's frame. In either case, the solution to the sub-problem is 3h. If you were asked to find the total time, add the first hour back in to get 4h.

Here are some graphs to help you visualize the different frames of reference and the movements of the trains overall:

I need a length in order to calculate this. I can't simply plug in the 45 km from above because that would be the time in which the second train got to the 45km mark, but the first train would have moved away from that point.

You just found here a very good way to solve the problem. But you're right, since the trains are constantly moving, you cannot just take the 45 km distance and use it.

When you take the train, you have probably already noticed that when it starts, it sometimes seems as if the platform is moving backwards. Or more likely, when your train is parked directly next to another one and one of them starts moving, you can't tell which one has actually started.

Choose a simple frame of reference. At the moment when the second train starts, if it is travelling at 60 km/h and the second at 45 km/h, it seems to be approaching the first at 60 - 45 = 15 km/h in its reference frame. Everything happens as if he had to travel 45 km at 15 km/h.

Your beginning of response was perfectly correct: the time needed for the first to catch up with the second is 45 (km/h) / 15 (km) = 3 hours.

This is not a visual explanation, but if you're trying to grasp the concept, it's this: What stays the same in the problem? Identifying this is the way to approach many physics problems.

Sometimes, a problem is looking for a sameness "before and after." That's often the approach to take for problems related to conserved quantities such as momentum or energy.

For this problem, it's not a before and after issue. Instead, you need to identify: What stayed the same between the two trains? And the answer, of course, is distance. Both trains traveled the same distance.

And of course you know that distance is the product of rate and time: $$d = rt$$.

Hopefully, from here, you may already see the approach. It's just algebra. In fact—since you are looking for a conceptual approach—I am half tempted to not even show the actual solution so that this stays as a conceptual answer and not a cookbook. But, for the sake of being complete, here it is.

If I use subscripts of 1 for train 1 and subscripts of 2 for train 2:

$$d_1 = r_1t_1$$ $$d_2 = r_2t_2$$

You know that both trains traveled the same distance, so $$d_1 = d_2$$

And substituting for $$d_1$$ and $$d_2$$:

$$r_1t_1 = r_2t_2$$

Now, you are given the speeds of both trains in the problem, and you are given the relationship between the times: train 1 traveled for one hour longer than train 2. So:

$$t_1 = t_2 + 1 hr$$

At this point you know 3 out of 4 of the values in: $$r_1t_1 = r_2t_2$$

Plugging in those values...

$$(45 km/hr)(t_2 + 1 hr) = (60 km/hr)(t_2)$$

Finally, to solve for $$t_2$$, since that is what the question is asking for:

Cancel units of km/hr from both sides of the equation:

$$(45)(t_2 + 1 hr) = (60)(t_2)$$

Divide both sides of the equation by 45:

$$t_2 + 1 hr = (4/3)(t_2)$$

Subtract $$t_2$$ from both sides:

$$1 hr = (1/3)t_2$$

Divide both sides by one-third:

$$3 hr = t_2$$.

Great question. In fact, for this question, there are many ways to solve it, all of which anyways end up with the same answer. I am going to explain in my way. First let's see what we know. Here, we have two trains, say train A and train B, train A moving at a speed of $$45\frac{km}{h}$$ and train B moving at a speed of $$60\frac{km}{h}$$. Now, train B starts moving $$1$$ hour after train A starts moving. So, $$1$$ hour after train A starts moving, train A is $$45$$ km ahead of train B. You want to know when they both will meet each other again. Usually, you would find the LCM of $$45$$ and $$60$$. The LCM of $$45$$ and $$60$$ is $$180$$. Meaning, both trains would meet each other after travelling $$180$$ km in total. If train A started at the starting point itself, it will take $$4$$ hours to travel $$180$$ km, and train B will take $$3$$ hours to travel $$180$$ km. In this question's case, train A is already $$1$$ hour ahead, meaning it would take train A, $$3$$ hours to travel $$180$$ km in total. Therefore, both train A and train B would take same time to travel $$180$$ km in total, in this case. And, you basically found the answer! Both trains will meet each other after 3 hours.

You want to visualize something that takes time to travel a distance. Its hard to imagine actual trains moving at actual speeds, so you would likely be better off with a substitute.

How about a Monopoly game board? Two trains, two pieces.

Train 1 is 45km/h. Train 2 is 60km/h. The difference is 15km/h. Luckily, all three are divisible by 15. That gives us Train 1 at 3km/h, Train 2 at 4km/h, and the difference at 1km/h.

Monopoly boards don't have kilometers of space to use, but you can use those reduced numbers to associate with number of spaces on the board.

Now each hour of time can be replaced with a "turn" in the game. On turn 1, Train 1 moves 3 spaces and Train 2 stays put. Turn two Train 1 moves 3 more spaces, Train 2 moves 4 spaces. Repeat until you land on the same space at the end of the turn.

In this manner, you're not calculating everything at once. Here you visually and physically calculate each hour as a turn until arriving at the compiled answer.

As the others have said, visualizing the problem is very helpful. What ever works for you is what you should use. If you still have trouble, try something else. Such as; laying a yard (meter) stick on the table. Use 2 coins, say a Dime and a Nickle. The nickle travels at 4.5 (inches/cm) per hour. The dime travels at 6. Move the nickle 4.5 to represent the first hour. (Add this to a table and mark it as 1 Hr), Move the nickle to 9 and the dime to 6 to represent the second hour (Add to a table as 2 Hr), etc.

Ultimately, since this is a physics class, you will need to translate this into equations.

Distance = Rate multiplied by Time (D=R*T). This equation you will use in your visualization, especially if you try time intervals other than 1 hour.

Now the "Math Speak":
The distance the first train travels (D1) is 45T.
The Distance the second train travels (D2) is 60
(T-1).
Obviously, When D1=D2, the second train has caught the first.
Or, in Math:
45T=60(T-1)

Please don't forget this step! One of the tenants of Physics is to understand how the world works and describe it mathematically!

Let $$t$$ be the time elapsed by the second train since it started to move.

The positions of the first train and the second train relative to the same starting point are $$x_1(t)$$ and $$x_2()$$ as follows.

$$\begin{cases} x_1(t)= S_\text{ref} + v_1 t\\ x_2(t)= v_2 t \end{cases}$$

where $$S_\text{ref}$$ is the initial relative position.

At $$t=T$$ the second train catches the first one. So they must be at the same position. \begin{align} x_1(T) &= x_2(T)\\ S_\text{ref}+v_1 T &= v_2 T\\ S_\text{ref} &= (v_2-v_1)T\\ T&= \frac{S_\text{ref}}{v_2-v_1} \end{align} If we denote $$v_2-v_1$$ as $$V_\text{ref}$$ then $$T= \frac{S_\text{ref}}{V_\text{ref}}$$

So we can say

The time needed by the second train to catch the first train is equal to the ratio of their initial spatial separation to relative speed.