What is the ideal timetable duration for a complete train journey?

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chwala
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Homework Statement
see attached
Relevant Equations
understanding of the ##v-t## graph
The question is below;
1639697693005.png
My approach;
and kindly note that i am comfortable in using the si units. Here i used,
##60##km/hr = ##16.66666##m/s.

For part (a),

using the ##v-t## graph,
##→s_1##=(##\frac {1}{2}##×##180×16.666)##+(##\frac {1}{2}##×##60##×##16.666##)
##=1500+500=2000##m
Total distance from Aytown to City is ##9000+7000=16000##metres
Therefore the remainder portion is ##16000-2000=14000##m. The time taken to cover this ##14000## m is given by;
##t##=##\frac{14000}{16.6666}##= ##840## seconds.
The timetable therefore should give an allowance of ##t=180+60+840=1080## seconds = ##18##
minutes.

For part (b),
using the ##v-t## graph,
Total distance from Aytown to Beeburg is ##9000##metres
##→s_1##=(##\frac {1}{2}##×##180×16.666)##+(##\frac {1}{2}##×##60##×##16.666##)
##=1500+500=2000##m
Therefore the remainder portion is ##9000-2000=7000##m. The time taken to cover this ##7000## m is given by;
##t##=##\frac{7000}{16.6666}##= ##420## seconds.

Total distance from Beeburg to City is ##7000##metres
##→s_1##=(##\frac {1}{2}##×##180×16.666)##+(##\frac {1}{2}##×##60##×##16.666##)
##=1500+500=2000##m
Therefore the remainder portion is ##7000-2000=5000##m. The time taken to cover this ##5000## m is given by;
##t##=##\frac{5000}{16.6666}##= ##300## seconds.
The timetable therefore should give an allowance of ##t=180+60+420+1+180+300+60=1201## seconds = ##20.016666##
minutes.
The time allocated should be ##21## minutes...bingo:cool:

any other approach guys...
 
Last edited:
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Hi @chwala ,

What is $$
s_1=(12×180×16.666)+(12×60×16.666)
$$in symbols ? And in words ?

[edit] check your units on$$
t=180+60+420+1+180+300+60=1201$$##\ ##
 
Last edited:
BvU said:
Hi @chwala ,

What is $$
s_1=(12×180×16.666)+(12×60×16.666)
$$in symbols ? And in words ?

[edit] check your units on$$
t=180+60+420+1+180+300+60=1201$$##\ ##
The units are just fine ...##1201## seconds, ##s_1## was the first portion of the ##v-t## graph...
 
##s_1## is something else !

What is the 1 second for ?

##\ ##
 
1639752348616.png


I hope this is clear for part (a)...
 
BvU said:
##s_1## is something else !

What is the 1 second for ?

##\ ##
This is for part (b) of the question, i guess i need to change the font color to distinguish between the two parts. Consider the original question they had indicated, ...at an intermediate station ##1## minute is allowed...
 
So ##s_1## is the distance needed to come up to speed and to decelerate to a stop. And to calculate it you use your favourite ##s = \frac 1 2 (v_{\rm start} + v_{\rm end})\,t\ ## which is fine.

1 second is not the same as one minute ! You want to add 60, not 1 for the stop.

Other than that, just fine. Bravo!

##\ ##
 
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BvU said:
So ##s_1## is the distance needed to come up to speed and to decelerate to a stop. And to calculate it you use your favourite ##s = \frac 1 2 (v_{\rm start} + v_{\rm end})\,t\ ## which is fine.

1 second is not the same as one minute ! You want to add 60, not 1 for the stop.

Other than that, just fine. Bravo!

##\ ##
That's right...yeah i should have converted the ##1## minute into seconds...let me post the correct solution after this post.
 
...for part (b)...

The timetable therefore should give an allowance of ##t=180+60+420+60+180+300+60=1260## seconds = ##21## minutes.
The time allocated should be ##21## minutes...bingo:cool:
 
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