Two Stretches, Five Halts, and a 60 km/h Demand
A 300 km mail route is divided into two equal stretches of 150 km each. On the first stretch the permitted speed is 50 km/h and the crew must make three scheduled halts of 6 minutes each. On the second stretch the permitted speed is 75 km/h with two scheduled halts of 6 minutes each. The timetable allots 5 hours for the run, and the control office has told the guard that the train must average 60 km/h over the whole 300 km, halts included, to hold its path. The guard points out that the first stretch is effectively fixed; only the second stretch can be changed, and only within the section's permitted speed.
1. At the stated speeds and halts, what is the train's average speed over the full 300 km?
Read the worked answer
Stretch 1 (150 km at 50 km/h)
- Running time = 150 / 50 = 3 h
- Halts = 3 × 6 min = 18 min = 0.30 h
- Block time = 3.30 h
Stretch 2 (150 km at 75 km/h)
- Running time = 150 / 75 = 2 h
- Halts = 2 × 6 min = 12 min = 0.20 h
- Block time = 2.20 h
Whole journey
- Total time = 3.30 + 2.20 = 5.50 h
- Average speed = total distance / total time = 300 / 5.50 = 54.55 km/h (approx.)
The common mistake is to average 50 and 75 to get 62.5 km/h. That is wrong twice over: the two stretches are equal in distance so their times must be weighted by time, and the halts must be added to the denominator, not ignored.
2. What speed would be needed on the second stretch to achieve the demanded 60 km/h overall? Is that realistic?
Read the worked answer
Target: 300 km at 60 km/h → total time = 300 / 60 = 5.00 h.
Time already consumed: Stretch 1 takes 3.30 h (3 h running + 0.30 h of halts), and that is fixed.
Time left for stretch 2: 5.00 − 3.30 = 1.70 h, which must also contain the 12 minutes (0.20 h) of halts.
- Available running time = 1.70 − 0.20 = 1.50 h
- Required speed = 150 / 1.50 = 100 km/h
Is it realistic? No. The section's permitted speed on that stretch is 75 km/h; 100 km/h is 33% above the limit and cannot be run without a special sanction, and the 75 km/h-limited stretch would only give 150/75 = 2 h, i.e. 0.5 h too slow.
So the demand is arithmetically possible but operationally illegal. The honest options are: cut halts, obtain an upgraded speed certificate, or renegotiate the path - not simply to instruct the driver to go faster.
3. If every halt on both stretches is cut to 4 minutes and the second stretch stays at 75 km/h, what average speed results?
Read the worked answer
Stretch 1
- Running = 3.00 h
- Halts = 3 × 4 min = 12 min = 0.20 h
- Block = 3.20 h
Stretch 2
- Running = 2.00 h
- Halts = 2 × 4 min = 8 min = 0.1333 h
- Block = 2.1333 h
Total = 3.20 + 2.1333 = 5.3333 h
- Average speed = 300 / 5.3333 = 56.25 km/h
Comparison: 54.55 km/h before, 56.25 km/h after - a gain of about 1.7 km/h from saving 10 minutes of halt time.
Judgement: even with all halts squeezed, the train is still about 3.75 km/h short of the 60 km/h demand (it would need 5.00 h against 5.33 h). Cutting halts alone cannot deliver the path; the control office has to accept a slower timing or the section speed must rise.