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

• HAF
Acceleration is constant. 4.The question is asking for the time it takes for the 4th wagon from the end to reach me. To do that we need to solve for t4 which is just t3 + 4.So guys I talked to my teacher and he clarified few important things. In summary, the first wagon passes next to me in 4 seconds. The n-th wagon will pass next to me in 4 seconds + 4.

## Homework Statement

Train starts to travel in a straight line at increasing speeds. The first wagon passes next to me in 4 seconds. In how many seconds will the n-th wagon pass next to me?

## Homework Equations

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

## The Attempt at a Solution

First of all we didn't learn sequences at school yet. That's why I try to solve it with these equations. Maybe the right question is "Is it solvable without knowing sequences?"

I tried to use the equation for velocity and compare then t1 with t2 with t3 and so on but there just so many unknowns and I can't figure anything out. Thank You I appreciate it.

This is similar to a free-fall scenario. In the first 4 seconds the distance is one train length, let's call that X. When do we get twice the distance (2X), which means the second wagon passed? You can express this with two equations and two unknowns or rearrange your first formula to use just a single equation.

HAF said:
at increasing speeds.
Is that all it says? Nothing about constant acceleration, or maybe constant power?

haruspex said:
Is that all it says? Nothing about constant acceleration, or maybe constant power?
I'm very sorry. I forgot to say that those wagons have same length.

haruspex said:
Is that all it says? Nothing about constant acceleration, or maybe constant power?
And acceleration is constant you are right.

HAF said:
And acceleration is constant you are right.

HAF said:

## Homework Statement

Train starts to travel in a straight line at increasing speeds. The first wagon passes next to me in 4 seconds. In how many seconds will the n-th wagon pass next to me?

Not difficult if you assume that "straight line" is not important, "increasing speeds" means "constant acceleration", and "The first wagon passes next to me in 4 seconds" means "At the start, I was standing at the front of the first wagon, which then passed me after 4 seconds", because you won't be able to solve it cleanly, otherwise.

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PeroK said:
Train starts to travel in a straight line at increasing speeds. The first wagon passes next to me in 4 seconds. Acceleration is constant and wagons have the same length. In how many seconds will the n-th wagon pass next to me?

HAF said:
Train starts to travel in a straight line at increasing speeds. The first wagon passes next to me in 4 seconds. Acceleration is constant and wagons have the same length. In how many seconds will the n-th wagon pass next to me?

It doesn't say how far away from you the train was when it started.

PeroK said:
It doesn't say how far away from you the train was when it started.
That's true. I was confused too but I take for granted what hmmm27 said.

hmmm27 said:
"increasing speeds" means "constant acceleration",
No, it just means the acceleration has the same sign as the velocity.

haruspex said:
No, it just means the acceleration has the same sign as the velocity.
But I know that acceleration is constant.

haruspex said:
No, it just means the acceleration has the same sign as the velocity.
True : edited "take for granted" to "assume" in the post. Of course, the actual assumption is that a nice clean answer is expected, rather than one laden with offsets. Both assumptions are odd enough to be worth mentioning in the submission, though (which could be taken as doing the student's homework for them, for which I apologize).

hmmm27 said:
True : edited "take for granted" to "assume" in the post. Of course, the actual assumption is that a nice clean answer is expected, rather than one laden with offsets. Both assumptions are odd enough to be worth mentioning in the submission, though.

To me "the wagon passes next to me" means the centre of the wagon, which is a also a solvable problem.

PeroK said:
To me "the wagon passes next to me" means the centre of the wagon, which is a also a solvable problem.

Well, if you insist... it could be a Wild West wagon train, in which case we'd want to know the distance between the arse end of a wagon and the horse pulling the next one, as well... and of course the length of the average horse.

PeroK said:
To me "the wagon passes next to me" means the centre of the wagon, which is a also a solvable problem.

To me that means that it takes 4 seconds for the whole wagon to pass (from front to back). I guess the OP will need to make assumptions about the actual meaning.

Ray Vickson said:
To me that means that it takes 4 seconds for the whole wagon to pass (from front to back). I guess the OP will need to make assumptions about the actual meaning.
So guys I talked to my teacher and he clarified few important things.

1. At the beginning I'm staying in front of the first wagon.
2. Wagons have the same length
3. Acceleration is constant.

Then you're all set

hmmm27 said:
Then you're all set [emoji2]
Yeah now I fully understand the problem. I found out the equation for every time by comparing time of first wagon, second wagon and third wagon.

Is there an easier way to find out the equation for time of the n-th wagon?

HAF said:
I found out the equation for every time by comparing time of first wagon, second wagon and third wagon.

Is there an easier way to find out the equation for time of the n-th wagon?

I'm entirely unsure how you'd find the time for the first, second, third wagon, without actually having a formula with an "n" in it. But, I'm lazy.

hmmm27 said:
I'm entirely unsure how you'd find the time for the first, second, third wagon, without actually having a formula with an "n" in it. But, I'm lazy.
I thought about it this way.
s1=s2 and then again s1=s3.

Ray Vickson said:
To me that means that it takes 4 seconds for the whole wagon to pass (from front to back). I guess the OP will need to make assumptions about the actual meaning.

It's also ambiguous in terms of whether you want the instantaneous time that (the end of) the nth wagon passes a given point; or, the length of time that the whole of the nth wagon takes to pass.

HAF said:
I thought about it this way.
s1=s2 and then again s1=s3.

Oh, well that's clear
Care to share ? in longhand... formulas and stuff.

PeroK
HAF said:
Yeah now I fully understand the problem. I found out the equation for every time by comparing time of first wagon, second wagon and third wagon.

Is there an easier way to find out the equation for time of the n-th wagon?
I suppose that you need some unit of length. Right?

Each wagon is the same length, so 'wagon' could be the length unit, and wagon/sec velocity, and wagon/sec2 the acceleration unit.

Then use kinematic equations

HAF
SammyS said:
I suppose that you need some unit of length. Right?

Each wagon is the same length, so 'wagon' could be the length unit, and wagon/sec velocity, and wagon/sec2 the acceleration unit.

Then use kinematic equations

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

HAF said:

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.

PeroK said:
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
PeroK 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.

HAF said:
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.

PeroK said:
The alternative is that the question is asking for the total time for n wagons to pass.
.
That's the way I interpreted the question. I do agree that the ambiguous wording is confusing.

PeroK said:
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.

HAF said:

Does this equation help?

HAF said:
s= vo.t + 1/2at^2

PeroK said:
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.

HAF said:
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.

PeroK said:
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.

HAF said:

##s = \frac12 at^2##

hmmm27