Teaser 2472: [Running from a train]
From The Sunday Times, 7th February 2010 [link]
George and Martha stood on the platform as the train approached at a steady 20 metres per second. Both platform and train were 100 metres long. The train was programmed to decelerate steadily as it reached the start of the platform. A boy slipped off the platform onto the track. He could run at 2 metres per second. “Run to your left — you’re nearer that end of the platform”, shouted Martha. “No, run to the other end, away from the train”, said George. Either action would just avert disaster.
How far ahead of the train did he fall?
This puzzle was originally published with no title.
Although this puzzle can be solved, the published solution was incorrect.
[teaser2472]
S2T2 
Jim Randell 8:08 am on 1 August 2025 Permalink |
I am assuming the train does not start any kind of emergency braking procedure, and so begins decelerating at the start of the platform, and ends (with the train stationary) aligned with the end of the platform.
In the first scenario the boy runs towards the train (which hasn’t begun to slow down yet). If he reaches the start of the platform (and dodges around it) after x seconds, just as the train arrives at the start of the platform, then he has run a distance of 2x metres, and the train has travelled 20x metres.
In the second scenario, the boy runs away from the train. After x seconds the train arrives at the start of the platform and starts to decelerate, such that its speed will be 0 when it arrives at the end of the platform (which is 100 m long).
The train starts at 20 m/s and after 100 m it reaches 0 m/s, the deceleration of the train is thus:
And the velocity of the train after time t from the start of the platform is:
So it takes the train 10 s to decelerate to 0 m/s.
We are interested in the time it takes for the train to decelerate to 2 m/s (after which the boy will outpace it, and get to the end of the platform before the train).
This happens at:
i.e. 9 seconds after the train reaches the start of the platform.
The distance along the platform travelled by the train in these 9 seconds is:
So there is 1 m of platform left.
And the boy has to run a distance of (99 − 2x) m in the time it takes the train to decelerate to 2 m/s.
The time taken for the boy to do this is:
And the time taken for the train to arrive at the same point is:
If disaster is just averted, these times are equal:
So the boy falls a distance 2x from the start of the platform = 40.5 m.
And at that time the train is 20x away from the start of the platform = 405 m.
Hence:
Solution: The boy falls at a distance of 445.5 m in front of the train.
Here is a time/distance graph of the situation:
The train enters from the top along the solid blue line (constant speed), and then as it passes the start of the platform it follows the dashed blue line (constant deceleration), until it stops at the end of the platform.
The scenarios for the boy are indicated by the red lines.
The upper red line is the path taken by the boy running toward the start of the platform (and the oncoming train), to arrive at the start at the same time as the train.
And the lower red line is the path take by the boy running towards the end of the platform (away from the train).
Zooming in on the final section we see that at 29.25 seconds the train catches up with the boy, as they are both going 2 m/s. The boy then outpaces the train to reach the end of the platform 0.5 seconds later, and the train finally stops at the end of the platform 0.5 seconds after that.
Whichever direction the boy chooses to run in, he has to start as soon as he lands, and so he doesn’t have to listen to advice from either Martha or George.
However, the published answer was 440 m.
This gives us:
Which means the boy fell 40 m from the start of the platform.
If we plot the distance/time graph associated with these values we see that although the train and the boy coincide at the end of the platform when the train is stationary, they also coincide 4 m before the end of the platform, when the boy is hit by the moving train. So this does not provide a viable solution to the puzzle.
Looking closer at the end we see at 28 seconds the train hits the boy, 2 seconds before they would both have arrived at the end of the platform.
At the point of impact the train would be travelling at 4 m/s (≈ 9 mph).
If the puzzle had been set so that, instead of trying to escape from the train, the boy was just trying to race the train to get to one end of the platform before the train does, then the published solution would work, and the boy and train would arrive simultaneously regardless of which end of the platform the boy chose to run to.
I couldn’t find any correction to the published solution in The Sunday Times digital archive.
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