Why does the other train seem to crawl, and how does a swimmer beat a river?

How to use this simulation

1. Start in the Two trains scene with the defaults: A at 20 m/s and B at 15 m/s in opposite directions. The relative speed reads 35 m/s, which is 126 km/h.
2. Switch the direction to Same direction. Watch the relative speed drop to the difference of the two speeds. Make B slower than A and the crossing time becomes "never".
3. Choose "From train A". Train A freezes on the screen and the milestones slide backwards. B now moves at exactly the relative velocity.
4. Open the River crossing scene. Leave the heading at zero and read the drift: aiming straight across does not land you straight across.
5. Pick the goal "Shortest path". The simulation tilts the swimmer upstream by sin θ = u/v and the landing point lines up with the green dashed line.
6. Raise the current above the swimmer's speed. The straight crossing becomes "not possible" and the readings show the least drift you can manage instead.

The train next door seems to roll backwards

You sit in a train at a station. The train on the next track starts moving, and for a moment you are sure that your own train is rolling backwards. Then you look at the platform, see it standing still, and the feeling collapses. What happened? Nothing moved except your point of view.

That little shock is the whole subject. Motion is never measured against nothing. It is measured against something: the ground, a train, the water of a river. Change that something and the same journey gets a different speed. This is what physics calls relative velocity.

Why does it matter for exams? Because two kinds of problems keep coming back. In one, two trains pass each other and you must find how long the crossing takes. In the other, a swimmer or boat crosses a river and you must find the time, the drift or the angle. Both look different. Both are the same subtraction.

Velocity depends on who is looking

Take two objects A and B, both measured against the ground. The velocity of A as seen from B is v(A relative to B) = v(A) − v(B). In words: take what A does, then take away what B does. If you are riding with B, B has no motion in your eyes, so everything else loses B's motion too.

On a straight line, directions become signs. Call right positive. A train going right at 20 m/s has +20; a train going left at 15 m/s has −15. Then A seen from B is 20 − (−15) = 35 m/s. The minus sign did the work for us, and that is why we write signs rather than guess "add" or "subtract".

Notice the symmetry. B seen from A is the exact opposite, −35 m/s. Same speed, reversed direction. That is natural: if the other train approaches you at 35 m/s, you approach it at 35 m/s too.

Words you must know

These terms appear in every question on the topic. Learn the meaning and the unit together.

Two trains: same direction or opposite

Here is the rule in one line. When the trains move in opposite directions the speeds add. When they move in the same direction the speeds subtract. That is it. It comes straight from the formula above: the signs of the two velocities are the same or different.

Why does it feel right? Drive past an oncoming bus on a highway and the bus seems to flash by. Overtake a slow truck going your way and you seem to creep past it for ages. The truck looks slow only because you are nearly as fast as it.

The table uses four pairs of speeds. Each number is worked out by code from the same formulas the simulation runs, so you can type any pair into the sliders and match the readout. Notice the third row: two trains at the same speed in the same direction have relative speed zero. One never gets past the other, and the simulation says "never".

With the defaults A = 20 m/s and B = 15 m/s, opposite directions, the relative speed is 20 + 15 = 35 m/s. Convert it for a feel: 35 m/s is 126 km/h, which is why two trains at a crossing look so frighteningly fast.

Crossing time: add the lengths

How long does a crossing take? Count from the moment the two fronts meet to the moment the two tails clear each other. In that time the trains must slide past each other by the sum of their lengths. So the time is the total length divided by the relative speed.

Why the sum of lengths? Picture standing on train A. Train B moves past you at the relative speed. It has completely passed when its whole length plus your own train's length has gone by. That is L₁ + L₂, one length for each train.

For the default pair the total length is 150 + 100 m and the relative speed is 35 m/s, so the crossing takes 7.14 s. Check it against the "Time to pass completely" readout. The shaded band between the tracks shows exactly the stretch where the two bodies are alongside one another.

The same formula covers a train crossing a platform, a bridge or a pole. A platform does not move, so the relative speed is just the train's own speed. A pole has no length, so only the train's length counts. One idea, three costumes.

Sit on train A: the world turns around

Choose the "From train A" viewpoint in the simulation. Train A stops on the screen. The milestones, which are fixed to the ground, now slide backwards at A's speed. Train B moves at the relative velocity.

This is exactly the trick that makes problems easy. Instead of juggling two moving trains, glue yourself to one of them and subtract its velocity from everything. Now one train is still and the other has a single speed. A two-body problem becomes a one-body problem.

Try it with the same-direction setting. In the ground view both trains move right, one a bit faster. In A's view only B moves, slowly, as if it were crawling up from behind. The arrow below B shows that velocity; the "Velocity of B in this view" readout gives its signed value.

The swimmer has two velocities at once

Now the river. A swimmer who can swim at v in still water jumps in, and the river itself flows at u. The water carries the swimmer along while the swimmer pushes through it. Seen from the bank, both movements happen together, so the velocities add as vectors.

Let the swimmer point at an angle θ measured from straight across toward upstream. Split the swimmer's own velocity into two parts. Across the river, v cos θ. Along the river, v sin θ pointing upstream. The current adds u downstream. So across the river the speed is v cos θ, and along the bank the speed is u − v sin θ.

The two directions never interfere. The time to cross depends only on the across component, because the current has no say in how fast you advance toward the far bank. The drift depends on how long you take and how fast you are carried along the bank.

The vector triangle at the bottom right of the simulation shows this: the green arrow is the swimmer in the water, the blue arrow the current, and the red arrow, from the start to the tip of the blue one, is the true velocity over the ground.

Shortest time or shortest path

These are two different goals and they ask for two different headings. For the least time, you want the biggest across component, so you point straight across, θ = 0. The time is W ÷ v whatever the current is. The cost is drift: the current carries you downstream.

For the shortest path you want to land exactly opposite the start, so the drift must be zero. That needs the current to be cancelled, u − v sin θ = 0, so sin θ = u ÷ v. You aim upstream. The price is time, because now only part of your effort goes across.

Take the default river: 100 m wide, swimmer 4 m/s, current 2 m/s. Straight across takes 25.0 s and lands 50.0 m downstream. The zero-drift heading is 30.0 degrees upstream and takes 28.9 s, because the usable speed across is √(4² − 2²) = 3.46 m/s.

The table below sweeps the heading angle for that river. The time only ever grows as you tilt upstream. The drift shrinks and reaches zero at the zero-drift angle; tilt further and you overshoot upstream, so the drift turns negative. The ground speed is the length of the red arrow.

When the swimmer is weaker than the river

What if the current is stronger than the swimmer, u > v? Then sin θ = u ÷ v would be bigger than 1, which is impossible. No heading can cancel the drift. The swimmer will always land downstream of the start. The "Straight-across angle" readout says "not possible" and the simulation shows the least drift instead.

The least drift is found by a neat trick: the resultant must be as close to straight across as the triangle allows. The best heading satisfies sin θ = v ÷ u, and the least drift is W√(u² − v²) ÷ v. It is more than the numbers suggest: in the last worked problem below the swimmer still lands more than the river's width downstream.

Try it: set the swimmer to 1 m/s and the current to 3 m/s, choose Shortest path, and see the red path leaning hard downstream while the swimmer points almost sideways against the flow.

Solved problems

Seven problems, from the easiest to the one that fools people. Each number is computed by the same formulas as the simulation, so you can reproduce all of them with the sliders.

Mistakes that cost marks

Look through this list before the exam. Every item has cost someone real marks.

• Forgetting to convert km/h to m/s when the lengths are in metres. Divide by 3.6, or keep everything in km.
• Adding speeds for a same-direction overtake. Same direction means subtract.
• Using only one train's length for a train-train crossing. Both lengths count.
• Using the sum of lengths when a train crosses a pole. A pole has no length.
• Mixing the axes in a river problem: time comes from the across component only.
• Writing sin θ = u ÷ v when v < u. That angle does not exist.
• Measuring θ from the bank instead of from the straight-across line. State your reference line in the answer.

Where else you meet relative velocity

Rain looks slanted from a moving bus even though it falls straight down: the rain's velocity relative to you is its own velocity minus the bus's. Pilots work with exactly the same triangle as the swimmer, with the wind in the place of the current, and tilt the nose into the wind to hold a course.

On the road, two cars heading toward each other at 15 m/s each have a closing speed of 30 m/s, which is why head-on collisions are so severe: the impact speed is the relative speed. The idea also reaches into astronomy and, later, into Einstein's relativity, where the simple subtraction has to be refined at very high speeds.

Frequently asked questions