Timetable Arithmetic for a 360 km Bus Service
A private bus operator runs a daily service between two towns that are 360 km apart on the highway. On a trial run the driver held a steady speed and completed the run in the scheduled time. On the return trial the next day, the driver raised the steady speed by 5 km/h and reached the destination 1 hour earlier than on the first day. The operator's timetable promises passengers an overall window of 6 hours for the journey, and this window already includes 40 minutes of halts at two intermediate stops; the buses are old and the manager doubts that any speed above 85 km/h is safe for them. Separately, a college group has asked for a faster service that completes the same 360 km within a 5-hour window, halts included. The manager wants the arithmetic settled before committing to either timetable.
1. Form a quadratic equation for the bus's usual steady speed and solve it. State clearly why one root is discarded.
Read the worked answer
Let the usual steady speed be x km/h, with x > 0.
- Time at usual speed = 360/x hours
- Time at increased speed = 360/(x + 5) hours
- Condition: 360/x − 360/(x + 5) = 1
Multiply through by x(x + 5): 360(x + 5) − 360x = x(x + 5) ⇒ 1800 = x² + 5x ⇒ x² + 5x − 1800 = 0.
Discriminant = 25 + 7200 = 7225 = 85², so x = (−5 ± 85)/2, giving x = 40 or x = −45.
The root −45 is rejected: a speed must be positive, and the step where we multiplied by x(x + 5) introduced no new physical case, so only x = 40 satisfies the original condition. Check: 360/40 − 360/45 = 9 − 8 = 1 hour ✓.
Usual speed = 40 km/h; usual running time = 9 hours.
2. Can the present 6-hour timetable promise be met at the usual speed? Support the answer with numbers, including the minimum speed the promise requires.
Read the worked answer
The 6-hour window already contains 40 minutes (2/3 hour) of halts, so the driving time allowed is 6 − 2/3 = 16/3 hours = 5 h 20 min.
Required speed = 360 ÷ (16/3) = 67.5 km/h.
At the usual 40 km/h the run alone takes 9 hours; adding the halts gives 9 h 40 min of journey time, more than 3.5 hours outside the promise. So the timetable is not met.
The speed needed, 67.5 km/h, is itself below the manager's 85 km/h safety ceiling, so the failure is a scheduling failure, not a safety constraint. The required increase is 67.5/40 = 1.6875, i.e. about 69% above the usual speed.
3. What steady speed would meet the 5-hour window, is it within the 85 km/h safety limit, and what percentage increase over the usual speed does it demand?
Read the worked answer
Driving time allowed = 5 − 2/3 = 13/3 hours ≈ 4 h 20 min.
Speed = 360 ÷ (13/3) = 1080/13 ≈ 83.08 km/h.
Safety check: 83.08 ≤ 85 ✓, but the margin is only about 1.9 km/h - any headwind, gradient or extra minute of halt pushes the schedule into unsafe territory.
Increase = 83.08 − 40 = 43.08 km/h; percentage increase = (43.08/40) × 100 ≈ 107.7%, i.e. the speed must more than double.
Judgement: technically possible, practically unsound. Trimming the halts is the cheaper lever: cutting 10 minutes of halt lowers the required speed to 360/4.5 = 80 km/h, buying about 3 km/h of margin at no fuel cost.