[pdurrant]

Originally Posted by jbjb

A bit late with this (only just saw the original post), but:

Spoiler:

Assuming the track stretches uniformly, the problem is isomorphic to one in which the track stays the same length, but his speed reduces each second. I.e. in the transformed problem, his speed starts at 10, then reduces to 10/2 after one second, 10/3 after the next and so on.

Hence, his distance travelled = 10 + 10/2 + 10/3 + 10/4 .....
which is 10 x (1 + 1/2 + 1/3 + 1/4 ....)

The progression in brackets can easiy be shown to sum to infinity, so (in the transformed domain) there is no limit to how far he can go, given time.

As the problems are isomorphic, the fact that he reaches the end in the transformed problem shows that he will reach the end in the original problem.

/JB

Yes, the right answer, and an interesting (and correct) way of calculating it. Congratulations!

You might want to search back for some of the previous puzzles. They all have the answers in spoiler tags, so you can still have a go at solving them.