The Problem:

The figure shows a map of the railway line from Aytown to City. The timetable is based on the assumption that the top speed of a train on this line is 60 km per hour; that it takes 3 minutes to reach this speed from rest, and 1 minute to bring the train to a stop, both at a constant rate; and that at an intermediate station 1 minute must be allowed to set down and pick up passengers. How long must the timetable allow for the whole journey (a) for trains which don’t stop at Beeburg, (b) for trains which do stop at Beeburg?

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Douglas Quadling Mechanics1 Exercise1D Q2

The figure shows a map of the railway line from Aytown to City. The timetable is based on the assumption that the top speed of a train on this line is 60 km per hour……

Douglas Quadling Mechanics1 Exercise1D Q2

The figure shows a map of the railway line from Aytown to City. The timetable is based on the assumption that the top speed of a train on this line is 60 km per hour

Problem Statement

The figure shows a map of the railway line from Aytown to City. The timetable is based on the assumption that:

• The top speed of a train on this line is 60 km/h.
• It takes 3 minutes to reach this speed from rest (acceleration).
• It takes 1 minute to bring the train to a stop (deceleration).
• At an intermediate station, 1 minute must be allowed to set down and pick up passengers.

The distances are:

• Aytown to Beeburg = 9 km
• Beeburg to City = 7 km

Questions:

1. How long must the timetable allow for the whole journey for trains that don’t stop at Beeburg?
2. How long must the timetable allow for the whole journey for trains that do stop at Beeburg?

Solution

Given:

• Top speed (v) = 60 km/h
• Time to accelerate (t_acc) = 3 minutes = 0.05 hours
• Time to decelerate (t_dec) = 1 minute = 0.0167 hours
• Time to stop at intermediate station (t_stop) = 1 minute = 0.0167 hours
• Distance from Aytown to Beeburg = 9 km
• Distance from Beeburg to City = 7 km
• Total distance from Aytown to City = 9 + 7 = 16 km

Step 1: Calculate Acceleration and Deceleration Distances

1. Acceleration Distance (s_acc):

sacc = (v * tacc) / 2
sacc = (60 * 0.05) / 2
sacc = 1.5 km

2. Deceleration Distance (s_dec):

sdec = (v * tdec) / 2
sdec = (60 * 0.0167) / 2
sdec = 0.5 km

Step 2: Calculate Time for Each Segment

1. Time to Accelerate (t_acc) = 3 minutes

2. Time to Decelerate (t_dec) = 1 minute

Step 3: Total Time for Trains That Don’t Stop at Beeburg

Distance at constant speed = Total distance - sacc - sdec
Distance at constant speed = 16 - 1.5 - 0.5 = 14 km
Time at constant speed = 14 / 60 = 0.2333 hours = 14 minutes
Total Time = tacc + tconst + tdec
Total Time = 3 + 14 + 1 = 18 minutes

Step 4: Total Time for Trains That Stop at Beeburg

1. Aytown to Beeburg:
   Distance at constant speed = 9 - 1.5 - 0.5 = 7 km
   Time at constant speed = 7 / 60 = 0.1167 hours = 7 minutes

2. Beeburg to City:
   Distance at constant speed = 7 - 1.5 - 0.5 = 5 km
   Time at constant speed = 5 / 60 = 0.0833 hours = 5 minutes

Total Time = tacc + tconst1 + tdec + tstop + tacc + tconst2 + tdec
Total Time = 3 + 7 + 1 + 1 + 3 + 5 + 1 = 21 minutes

Final Answers:

1. Trains that don’t stop at Beeburg: The timetable must allow 18 minutes.
2. Trains that stop at Beeburg: The timetable must allow 21 minutes.