Like Ships Passing (February 2026)

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So Vessel A, Ironweed travels 500 yds and meets Vessel B Lake Dog, then travels another 300 yds to reach Dock B. It then leaves Dock B and travels 300 yds before meeting Lake Dog again.

So Ironweed has travelled at total of 1100yds.

Vessel B Lake Dog travelled 300yds when it met Ironweed for the first time, it then continued to Dock A 500 yds away. It then left Dock A and met Ironweed again 300 yds from Dock B.

So lake Dog travelled 1300 yds in the same time as Ironweed covered 1100yds.
Difference is Lake Dog is 18.182% faster than Ironweed

Distance between the docks = 800 yds.

"The vessels then leave docks for return trips at the same time again"

So Rixter's answer is correct.

Exactly, both leave the docks at the same time and Lake Dog covers 500 yds. in the same time as Ironweed takes to cover 300 yds, so Lake Dog is faster than Ironweed.

Yes, Lake Dog is faster than Ironweed… on her Eastbound trip across the lake.

Lake Dog is slower than Ironweed on her Westbound trip across the lake.

Both ships: Depart Dock A, bound for Dock B (West shore to East shore) will intercept at 500 yds from Dock A.

Both ships: Depart Dock B, bound for Dock A (East shore to West shore) will intercept at 300 yds from Dock B.

Being sister ships they should be nearly identical construction and capabilities. So, both vessels probably have screws set the same, and a Westerly wind is blowing.

I also believe Rixter is correct in #3.

Rixter's second answer that it is 800 yds and going from A to B is 5/3 the speed of going from B to A is nice and it makes sense; however, the real answer is that we don't know the distance across the lake because the question doesn't say that it took the same total amount of time for both ships to finish their journey to the other side of the lake.

In reality there are too many unknowns. Since the speed of one boat is not constant going from one side of the lake to the other, who is to say that the speed of the boat is not changing both times. Maybe the distance across the lake is 1500 yds, and Ironwood is traveling at 1/3 the speed of Lake Dog the first trip going from A to B and then the second trip Ironwood is traveling at 1/5 the speed of Lake Dog going from B to A.

I agree with you that there are a lot of things unspecified and an infinite number of possibilities. I am assuming that the ships travel in a straight line and constant (although possibly different) speed for each of the four crossings. This narrows it down to a manageable number of possibilities.

There are two meetings as the ships pass, 500 yards from side A and 300 yards from side B. These could be the same spot, meaning the width of the lake is 800 yards. This would be the simplest case, needing only 2 speeds. The ship traveling from A to B has 5/3 the speed of the ship moving in the opposite direction.

I wrote a little MATLAB code to model different lake sizes with time-position plots:

SHORE A

SHORE B

Simplest case, 800 yards lake width. Meeting place 500 yards from side A and 300 yards from Side B.

Other lake sizes are possible requiring 4 different speeds.

SHORE A

SHORE B

600 yards width of lake. Note that 500 width is the absolute minimum where the meeting point would be on Shore B

SHORE A

SHORE B

Okay… yeah… maybe.

I believe the challenge question has a tricky word; differential. Why this word? Why not ‘ratio’?

I am prone to analysis paralysis, thus I am interpreting ‘differential’ here as ‘what makes them different?’, perhaps restated as ‘why are they different?’.

This interpretation also indicates that some sort of a pattern or relationship is observable.

Your #3 and spacetime graph in #10 show a pattern, and I offered up Prevailing Westerlies as explaining the pattern. Could also be lake currents, I guess.

I dunno, am I obsessing over an AI autofill event? Sometimes a cigar is, after all, just a cigar.

‘why are they different?’

The reason the speeds are different is that the crossover points are 500 yards from A and 300 yards from B. If you made them 400 yards from both shores, it would, of course, be the same, by symmetry.

I believe it should be:

The reason the speeds are different is that the crossover points are 500 yards from A and 300 yards from B is because the vessel speeds are different, easterly versus westerly heading.

If you made them 400 yards from both shores (the boats) the same actual speed, east and west, it would, of course, be the same, by symmetry.

I think we are saying about the same thing.

The meeting location (500 yd from A and 300 yd from B) was specified in the problem. For both ships to start simultaneously and meet simultaneously requires the ship moving west to east, the longer distance, to travel faster. (Ironseed on the first lap and Lake Dog on the return lap.)

It's the meeting location that requires different speeds for the two ships.

Exactly what CoPilot calculated when I just asked the question !

Let us run a number through the solution in post #8.

Let Ironweed speed be 100 yd/min.
Thus; Lake Dog speed at Ironweed plus 40% is 140 yd/min.

So: Ironweed trip A→B, given intercept 500 yds from A @ 100 yd/min = 5 min.
Thus; Lake Dog, 5 min @ 140 yd/min =700 yds. 500 yds + 700 yds = 1,200 yds.

So: Ironweed trip B→A, given intercept 300 yds from B @ 100 yd/min = 3 min.
Thus; Lake Dog, 3 min @ 140 yd/min = 420 yds. 300 yds + 420 yds ≠ 1,200 yds.

Conclusion: Solution in #8 does not satisfy conditions given.

Or… where am I going wrong?

Here is what I think the answer is:

Given no other information, the Lake is 800 yd wide. The dotted line is 500 yd from shore A and 300 yd from shore B.

Both crossover points are on this line. Ironweed starts at the top left corner and Lake Dog in the bottom left. From the slope of the path in the "spacetime diagram", Lake Dog travels at 3/5 the speed of Ironweed on the first pass and Ironweed travels at 3/5 the speed of Lake Dog on the return.

The REAL question is, why did this only show up in my email box on 2/17, when there are answers posted from two weeks ago?

The correct answer is that my email took the slow boat.