# Moving train problem

1. May 9, 2010

### Logistix

NOTE: This is no homework, just me having fun with physics, maths, logics etc. Do not take this problem as priority, there are surely more serious and urgent problems out there!

1. The problem statement, all variables and given/known data
A passenger standing beside the railway noticed that the first wagon of the electric train, decelerating linearly, passed beside him after 4 s, and the second wagon after 5 s. The first wagon stopped at 75 m in front of the passenger. Calculate the acceleration of the train.

t1 = 4 s
t2 = 5 s
s1 = 75 m

2. Relevant equations
s = v0*t - a*t^2/2
v = v0 - at

3. The attempt at a solution

Assuming the second wagon was the last and that it stopped at the same time as the first and that both are of same length, it is to say that the length of both wagons is 75 m. This also means it took the train 75 m and 9 total seconds to decelerate to 0. Now I wrote:
75 = 9*v0 - (81*a)/2
Since we don't know v0, we insert it from the following equation:
0 = v0 - 9a
v0 = 9a
--> 75 = 9*9a - (81*a)/2
75 = 81a - (81a)/2
after solving: a = -1.85 m/s^2
Then we find v0:
0 = v0 - 16.65
v0 = 16.61 m/s

As I do not have the solutions and the page from where I downloaded the assignments (in Croatian) didn't even offer them, I'm just asking from your opinion and critics and eventually the right solution.
I also tried this by not including v0 and using a different concept, however without it even more weird solutions came up. Any help is appreciated.

2. May 9, 2010

### kuruman

Hi Logistix, welcome to PF.

You cannot assume that each wagon is 75 m long. That's a pretty long wagon if you think about it. You have three unknowns, the length of each wagon L, the acceleration a and the initial speed v0. Therefore, you need three equations. You have found two of them

L = v0*4 - a*42/2
2*L = v0*9 - a*92/2

Can you think of a third one using the fact that the front wagon travels 75 m before it stops?

3. May 10, 2010

### Logistix

Right, that came to my mind in the beginning but I didn't want to work with equation systems, but it's all ok. I'd add the equation:

75 = v0*t - a*t^2/2
t = total time which we don't know, nor do we know the a so I replace:
75 = v0*t - v0*t/2 (a = v-v0/t -> in this case, v=0, so it can be simplified to a=-v0/t, in s form: s=at^2/2->s=v0/t*t^2->s=v0t)
Now t is a new unknown so I replace
0 = v0 - at We replace the t with the one in the previous expression:
150 = 2v0t-v0t
150 = v0t
t = 150/v0 -> 0 = v0 - 150/v0*a /*v0
0 = v0^2 - 150a
v0^2 = 150a

So the system is:
l = 4v0 - 16v0/2
2l = 9v0 - 81v0/2
v0^2 = 150a

After solving, which I did rather quickly due to some other obligations, I got a = 0.24 m/s^2.