Problem solved!
A problem (of the classical train variety) came my way, and I thought it would be good to share my thoughts about how I solved it on here as well as on Twitter:
Someone asked me for help with a howfastisthetrain question, and I thought it'd be interesting to document my thinking process as I solved it. All aboard! (1/) pic.twitter.com/QTJRc1yqWY
— Colin Beveridge (@icecolbeveridge) June 5, 2021

Read through the question to get a sense of it.
 What am I looking for?
 The (expected) speed of the train
 What would help with that?
 The distance we’re travelling – we’re given that, it’s 60 miles.
 The time it takes – we’re not given this.
 What are we given?
 That we lose two minutes over the first forty miles
 That going 12mph faster means we catch up over the last twenty miles.
 What variables might we use?
 I’m going to want something for the train’s initial speed – maybe $s_i$
 Probably something for the train’s expected speed – let’s go with $s_e$
 Total time taken, $T$.
 Restate the problem:
 Going from the first point to the second point (40 miles) takes two minutes longer at $s_i$ than it would at $s_e$
 Going from the second point to the last point (20 miles) takes four minutes less at $(s_i+12)$ than it would at $s_e$
 Think about units:
 Given the problem, the ideal units are probably mph, miles and hours.
 Two minutes is 1/30 of an hour; four minutes is 1/15 of an hour.
 Write some equations:
 Hold on, I need to know how speed and time relate.
 How do things relate?
 At constant speed, speed &time; time = distance
 time = distance / speed
 NOW write some equations:
 $\frac{40}{s_i}  \frac{40}{s_e} = \frac{1}{30}$
 $\frac{20}{s_e}  \frac{20}{s_i + 12} = \frac{1}{15}$
 What do I notice?
 It’s not nice having the variables on the bottom.
 It’s easy to eliminate the $s_e$ by doubling the bottom equation.
 It might be better to work in 30ths.
 Simplify:
 $\frac{40}{s_e} = \frac{40}{s_i}  \frac{1}{30}$
 $\frac{40}{s_e} = \frac{40}{s_i + 12} + \frac{4}{30}$
 So $\frac{40}{s_i}  \frac{1}{30} = \frac{40}{s_i + 12} + \frac{4}{30}$
 Or $\frac{40}{s_i}  \frac{40}{s_i + 12} = \frac{5}{30} = \frac{1}{6}$.
 Thought:
 It’s really $s_e$ I want, so it might have been better to eliminate $s_i$
 It’s easy enough to recover $s_e$ once I know $s_i$, so I’ll carry on
 Combine:
 $\frac{40(s_i + 12)  40 s_i} {s_i(s_i + 12)} = \frac{1}{6}$
 Simplify and crossmultiply $40 \times 12 = \frac{1}{6} s_i (s_ i + 12)$
 Solve for $s_i$:
 It’s a quadratic, should be straightforward
 $40 \times 72 = s_i (s_i + 12)$
 $48 \times 60 = s_i (s_i + 12)$ so $s_i = 48$
 Solve for $s_e$:
 $\frac{40}{s_e} = \frac{40}{s_i}  \frac{1}{30}$
 $\frac{40}{s_e} = \frac{40}{48}  \frac{1}{30}$
 $\dots = \frac{5}{6}  \frac{1}{30}$
 $\dots = \frac{24/30} or \frac{4}{5}$
 $s_e = 50$mph
 Check:
 The train travels 40m at 48mph, taking 50 minutes
 It then travels 20m at 60mph, taking 20 minutes
 So it’s travelled 60m in 70 minutes; it was scheduled to take 72.
 60m in 72 minutes is 50mph.
 The scheduled speed was 50mph.