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

[chwala]

Homework Statement

see attached

Relevant Equations

understanding of the $v-t$ graph

The question is below;

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

any other approach guys...

 

[BvU]

Hi @chwala ,

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

[edit] check your units on$$
t=180+60+420+1+180+300+60=1201$$$\$

 

[chwala]

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...

 

[BvU]

$s_1$ is something else !

What is the 1 second for ?

$\$

 

[chwala]

I hope this is clear for part (a)...

 

[chwala]

$s_1$ 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...

 

[BvU]

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!

$\$

 

[chwala]

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.

 

[chwala]

...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