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:

• 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 $s_e$
• 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 cross-multiply $40 \times 12 = \frac{1}{6} s_i (s_ i + 12)$
• 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.