2011 Headway calculation

[PJW]

I haven't done the maths, but yes I THINK you have understood although I'd have expressed a bit differently (I think you didn't write quite what you meant to)

The headway time for a non stopping train following a stopping train can be calculated by adding:
1. the headway time for a non-stopping train following a non-stopping train,
2. the difference between the time taken by a stopping train and a non-stopping train (i.e. the time taken for a stopping train to decelerate, dwell and re-accelerate to the headway speed, minus the time a non-stopping train would take to cover the same distance).

Ht (f-s) = Ht (f-f) + [t(s) - time taken for t(f) to cover the same distance]
where f=fast (non stop) and s=slow (stopping)

Actually this is a bit of a "cheat" method since it purely relies on consideration of Newton's Laws and does not consider the signalling for the situation prior to the fast train catching up the slow to be at minimum no-stop headway distance. It's ok for this question, but not for a calculation of "slow following slow", when the aspects seen by the 2nd train are relevant.

Thanks PJW,

So let me see if I have understood this in my own words. I will write it up later to see if I really have got it.

The headway time for a non stopping train following a stopping train can be calculated by adding the difference from the headway time for a stopping train following a stopping train, minus the time a non stopping train would take to cover the same distance.

Ht (ns f s) = Ht (ns) + [Ht(s f s) - time taken for Ht(ns) to cover the same distance]

It looks more understandable on my sketch pad!
And it only appears to add an extra 110 seconds to the headway time, which would fit in with the question requirements. Making 3mins 10secs headway time.

Regards
KJB