# Homework Help: Calculate the time for train

1. Aug 25, 2011

### tavo0116

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between stops. The more stops there are, the slower the train's average speed. To get an idea of this problem, calculate the time it takes a train to make a 6.0 trip in two situations. (Assume that at each station the train accelerates at a rate of 1.1 until it reaches 95 , then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at -2.0 . Assume it stops at each intermediate station for 23.0 .)

a.Calculate the time it takes a train to make a 6.0 trip if the stations at which the trains must stop are 1.2 apart (a total of 6 stations, including those at the ends).

b.Calculate the time it takes a train to make a 6.0 trip if the stations are 3.0 apart (3 stations total).

What kind of equation I should us?

Thank you.

2. Aug 25, 2011

### PeterO

Are all these values in metres and m/s and m/s^2 ?

You will use the normal equations, but the trip will be broken into many many parts.
Initial acceleration
Cruise towards 1st stop
Slow for first stop
wait at station
accelerate away
repeat many times: accelerate, cruise, slow, wait, accelerate, cruise, slow, wait etc

Fortunately, each acceleration and deceleration involve the same time - distance so you don't have to recalculate too often.

3. Aug 25, 2011

### tavo0116

Sorry for not entering the units.

Here is the one with the correct units.

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between stops. The more stops there are, the slower the train's average speed. To get an idea of this problem, calculate the time it takes a train to make a 6.0 km trip in two situations. (Assume that at each station the train accelerates at a rate of 1.1 m/s^2 until it reaches 95 km/h , then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at -2.0m/s^2 . Assume it stops at each intermediate station for 23.0 s .)

a. Calculate the time it takes a train to make a 6.0 km trip if the stations at which the trains must stop are 1.2 km apart (a total of 6 stations, including those at the ends).

b. Calculate the time it takes a train to make a 6.0 km trip if the stations are 3.0 km apart (3 stations total).

I know you've already given me the hints. But I'm really clueless on how I should start.

4. Aug 25, 2011

### PeterO

Start by converting 95 km/h to m/s, then calculate how long - both time and distance - it takes to accelerate from rest to that speed [at 1.1 m/s^2].
Also calculate the time and distance involved in stopping from 95 km/h at the stated acceleration [-2.0 m/s^2]

5. Aug 25, 2011

### tavo0116

So as you stated, I will start by converting all km/h to m/s.
Which then I will get:

95 km/h = 26.39 m/s
6 km = 6000 m
1.2 km = 1200 m
3 km = 3000 m

I will then have the velocity (95 km/h) and acceleration (1.1 m/s^2). Which equation should I use to find distance and the time?

6. Aug 25, 2011

### tavo0116

Not sure if this is correct.
I will need to use the formula: V = u + at, where,
V = final velocity
u = initial velocity (0 m/s)
a = acceleration
t = time

When everything's plugged in, I will get the following,
26.39 m/s = 0 m/s x (1.1 m/s^2)(t)
Which then I will get 23.99 seconds as the solution?

7. Aug 25, 2011

### PeterO

That looks OK, so each time the train speeds up it takes 24 seconds. How many times does each train speed up?

Also calculate the distance covered during the acceleration. Since you now know v,u,a and t you can use any one of the other motion equations.

Also calculate time and distance while slowing.

8. Aug 25, 2011

### tavo0116

Can I use the d = v x t formula to find the distance?

9. Aug 25, 2011

### PeterO

NO because the train is not travelling at constant speed during these acceleration phases.

10. Aug 25, 2011

### tavo0116

Which equation should I use?

Thank you.

11. Aug 25, 2011

### PeterO

There are 5 equations of motion for movement under constant acceleration.

v = u + at is one of them

can you list any 2 of the other 4?

12. Aug 25, 2011

### tavo0116

I remember one..
Is it

d = Vt + (1/2)(a)(t^2) ?

13. Aug 25, 2011

### PeterO

That will work, you know all the variables to substitute - though given you said v = u + at earlier, I would have preferred to see either

d = ut + (1/2)(a)(t^2)

or

d = vt - (1/2)(a)(t^2)

14. Aug 25, 2011

### tavo0116

I have solved the distance for the acceleration with the formula I posted above,
which gives me 317 m.
For the deceleration, the initial velocity is 26.39 m/s, right? And the final velocity is 0 m/s?
If that is correct, I would get 174 m , and 13.1995 s.

So for question a, the total distance is 6000 m, and 1.2 km apart for all stations, and 23 seconds waiting time,

23 x 6 stations = 138 seconds.

I'm stuck there again..

15. Aug 25, 2011

### PeterO

Draw a long horizontal line across the page.
Put a dot at each end - representing the first and last station.
Now put 4 more dots evenly spaced in the middle - representing the other stations

Immediately after the first station, shade a small section and label it a
Immediately before the second station, shade a small section and label it d
label the section between as c. label the intermediate stations S.

That is "a" for accelerate; "c" for constant speed; "d" for decelerate.

You can label each of the other gaps between stations the same way.

Now you know how long [distance] each acceleration section is, and how long each deceleration section is. From that you will know how long each constant speed section is.

Using d = v*t you can calculate how long [time] it takes to cover that distance.

Now the total trip will be a,c,d,s,a,c,d,s, ... etc
each "a" is 24 seconds long - you already calculated that
each "d" is 13.2 seconds long - you have calculated that
each "S" is 23 seconds long - that was given
each "c" is how ever long you calculate.

Add them all up and that is how long it takes to get from the first Station to the last.

When you have only 1 station along the way, there are fewer bits to sum, and the "c" section takes longer.

16. Aug 25, 2011

### PeterO

WARNING: REFRESH!! I corrected some key spelling errors!!!!

17. Aug 25, 2011

### PeterO

Another way to consider the problem!

Each time you stop at a station, it take 13.2 seconds to pull up, 23 seconds to wait then 24 seconds to accelerate away - a total of 60.2 seconds [I am assuming you time - distance for pulling up is correct.]
In that 60.2 seconds you travel only 174 + 317 = 491m

If instead the train just proceeded through at 95 km/h [ 26.39 m/s] it would only take 18.6 seconds. So every time you stop you "waste" 41.6 seconds.

18. Aug 25, 2011

### tavo0116

So to calculate the time that the train travels at the constant speed (26.39 m/s),
240 m = 26.39 m/s x t
Which then I will get 9.1 s?

19. Aug 26, 2011

### PeterO

Where did 240 m come from??

Distance between stations 1200 m

Distance covered while accelerating - apparently 317 m
Distance covered while slowing down - apparently 174 m

So that leaves more than 240 m for cruising ???

You have used the right formula.