# In how many seconds will the n-th wagon pass next to me?

PeroK
Homework Helper
Gold Member

How can I insert time(n) into those kinematic equations?

I would like to solve this problem as fast as it could be but comparing time(1) and time(2) and etc. Isn't fast enough.

I hope you won't get angry because of my question.

Thank You
One of the problems is that you haven't shown us any of your work yet. We've no idea how you are trying to solve this problem, what your algebra skills are like and how you use the kinematic equations.

For example, suppose you were asked to find the speed of the train after the nth wagon has passed, then I would do something like:

Let ##L## be the length of each wagon.

We have the kinematic equation: ##v^2 - u^2 = 2as##, which in this case (##u = 0##) reduces to ##v^2 = 2as##, where ##a## is the constant acceleration of the train.

Let ##v_n## be the speed of the train after ##n## wagons have passed. Then we have:

##v_n^2 = 2a(nL)##

And, suppose you were given that the speed of the train after the first wagon was ##v_1 = 2m/s##, and the length of each wagon is ##16m##. Note: this is a different problem from your one. Then:

##v_1^2 = 2aL##

##a = \frac{v_1^2}{2L} = \frac{4m^2/s^2}{32m} = 0.125 ms^{-2}##

And, finally,

##v_n^2 = 2(0.125 ms^{-2})n(16m) = (4n)m^2s^{-2}##

##v_n = (2\sqrt{n}) ms^{-1}##

That's not necessarily the quickest way to solve the problem, but I wanted to illustrate the way you should be approaching these problems. How to introduce variables, and use and solve the kinematic equations.

One of the problems is that you haven't shown us any of your work yet. We've no idea how you are trying to solve this problem, what your algebra skills are like and how you use the kinematic equations.
This is how I worked.

As you said
Let LL be the length of each wagon.
we know that L1 = L2 = Ln
t1 = 4 s

at the beginning we know that L1 = ½.a.t12 = L2 = L3
then I wrote equation L2 = v1.t2 + ½.a.t22 NOTE: v1 = a.t1
and L3 = v2.t3 + ½.a.t32 NOTE: v2 = a.t1 + a.t2

t2 = 4.(√2 - 1)
t3= 4.(√3 - √2)

and that's how I found out that tn = t1.[√n - √(n-1)]

But this way is very long i guess and I have no idea how to solve it faster.

PeroK
Homework Helper
Gold Member
This is how I worked.

As you said
we know that L1 = L2 = Ln
t1 = 4 s

at the beginning we know that L1 = ½.a.t12 = L2 = L3
then I wrote equation L2 = v1.t2 + ½.a.t22 NOTE: v1 = a.t1
and L3 = v2.t3 + ½.a.t32 NOTE: v2 = a.t1 + a.t2

t2 = 4.(√2 - 1)
t3= 4.(√3 - √2)

and that's how I found out that tn = t1.[√n - √(n-1)]

But this way is very long i guess and I have no idea how to solve it faster.
Okay, so you have interpreted the question as asking for the time that the nth wagon takes to pass. As I pointed out in post #22 the question was also ambiguous on this point. The alternative is that the question is asking for the total time for n wagons to pass.

If we let ##T_n## be the total time for n wagons to pass, then we have ##t_n = T_n - T_{n-1}##. And that should give you perhaps a quicker way.

SammyS
Staff Emeritus
Homework Helper
Gold Member
The alternative is that the question is asking for the total time for n wagons to pass.
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That's the way I interpreted the question. I do agree that the ambiguous wording is confusing.

Okay, so you have interpreted the question as asking for the time that the nth wagon takes to pass. As I pointed out in post #22 the question was also ambiguous on this point. The alternative is that the question is asking for the total time for n wagons to pass.

If we let ##T_n## be the total time for n wagons to pass, then we have ##t_n = T_n - T_{n-1}##. And that should give you perhaps a quicker way.

PeroK
Homework Helper
Gold Member
Does this equation help?

s= vo.t + 1/2at^2

Does this equation help?
What I did this time is this.
vn = vn-1 .atn
which means a=(vn -vn-1) / tn

and that's what I put in equation s=vn-1.tn + 1/2.a.tn2

what I got is this tn = 2s/(vn + vn-1)

Can this somehow help me? I'm really sorry for my stupidness but I can't figure it still out.

PeroK
Homework Helper
Gold Member
What I did this time is this.
vn = vn-1 .atn
which means a=(vn -vn-1) / tn

and that's what I put in equation s=vn-1.tn + 1/2.a.tn2

what I got is this tn = 2s/(vn + vn-1)

Can this somehow help me? I'm really sorry for my stupidness but I can't figure it still out.
It's not a good idea to consider intermediate speeds. That is where the complications arise. ##v_0 = 0## for the motion of the train.

It's not a good idea to consider intermediate speeds. That is where the complications arise. ##v_0 = 0## for the motion of the train.

PeroK
Homework Helper
Gold Member
##s = \frac12 at^2##

• hmmm27
##s = \frac12 at^2##
Nope. I can't figure anything out. I will make a break for a while and then I will try it again because now I can't do it.

Thank You you for everything. I really appreciate it.

mfb
Mentor
Plug in L for the first wagon.
Plug in nL for the first n wagons and solve for the new time. You should get ##t \sqrt n##. From there you can take the difference between adjacent wagons.

Plug in L for the first wagon.
Plug in nL for the first n wagons and solve for the new time. You should get ##t \sqrt n##. From there you can take the difference between adjacent wagons.
Plug it where? In s=1/2a.t^2?

Plug in L for the first wagon.
Plug in nL for the first n wagons and solve for the new time. You should get ##t \sqrt n##. From there you can take the difference between adjacent wagons.
Thank You guys! I finally got it!

DaveC426913
Gold Member
So guys I talked to my teacher and he clarified few important things.

1. At the beginning I'm standing in front of the first wagon. • 