Mechanics question ??????????
Aytown---------(9 km)-----------Beeburg-------(7km)-------city.
The figure shows a map of the branch line from Aytown to City. The timetable is based on the assumption that the top speed of a train is 60km 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 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? answer should be 18 min
b) for trains which do stop at Beeburg ? answer should be 21 min
The figure shows a map of the branch line from Aytown to City. The timetable is based on the assumption that the top speed of a train is 60km 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 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? answer should be 18 min
b) for trains which do stop at Beeburg ? answer should be 21 min
Original post by rawanaimat
Aytown---------(9 km)-----------Beeburg-------(7km)-------city.
The figure shows a map of the branch line from Aytown to City. The timetable is based on the assumption that the top speed of a train is 60km 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 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? answer should be 18 min
b) for trains which do stop at Beeburg ? answer should be 21 min
The figure shows a map of the branch line from Aytown to City. The timetable is based on the assumption that the top speed of a train is 60km 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 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? answer should be 18 min
b) for trains which do stop at Beeburg ? answer should be 21 min
What have you tried?
Original post by Kevin De Bruyne
What have you tried?
Spoiler
well, first let's split it up into chunks.
it takes 3 min to reach 60 km/h, we need to convert this into standard units (pref minutes, and the distance can stay in km),
then use s=0.5(u+v)t to work out the distance over which we have had this constant acceleration (s1)
Then you can work out the distance to bring the train to a stand still with the same equation (s2)
Then using v=s/t, and by subtracting the distance to start and stop from the distance between the stations (s(at 60km/h) = s(total) - s1 + s2), work out the time it takes to travel the distance at 60 km/h
Then add up the time and bingo, you've got the answer.
it takes 3 min to reach 60 km/h, we need to convert this into standard units (pref minutes, and the distance can stay in km),
then use s=0.5(u+v)t to work out the distance over which we have had this constant acceleration (s1)
Then you can work out the distance to bring the train to a stand still with the same equation (s2)
Then using v=s/t, and by subtracting the distance to start and stop from the distance between the stations (s(at 60km/h) = s(total) - s1 + s2), work out the time it takes to travel the distance at 60 km/h
Then add up the time and bingo, you've got the answer.
(edited 4 years ago)
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