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

[HAF]

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.

 

[mfb]

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.

 

[haruspex]

at increasing speeds.

Is that all it says? Nothing about constant acceleration, or maybe constant power?

 

[HAF]

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.

 

[HAF]

Is that all it says? Nothing about constant acceleration, or maybe constant power?

And acceleration is constant you are right.

 

[PeroK]

And acceleration is constant you are right.

Please post the complete question.

 

[hmmm27]

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.

 

[HAF]

Please post the complete question.

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?

 

[PeroK]

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.

 

[HAF]

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.

 

[haruspex]

"increasing speeds" means "constant acceleration",

No, it just means the acceleration has the same sign as the velocity.

 

[HAF]

No, it just means the acceleration has the same sign as the velocity.

But I know that acceleration is constant.

 

[hmmm27]

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).

 

[PeroK]

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.

 

[hmmm27]

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.

 

[Ray Vickson]

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.

 

[HAF]

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.

 

[hmmm27]

Then you're all set

 

[HAF]

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?

 

[hmmm27]

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.

 

[HAF]

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.

 

[PeroK]

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.

 

[hmmm27]

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.

 

[SammyS]

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]

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

May I ask something please?

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

 

[PeroK]

May I ask something please?

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.

 

[HAF]

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 = L_n
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 t_n = t₁.[√n - √(n-1)]

But this way is very long i guess and I have no idea how to solve it faster.
Thank You for your help.

 

[PeroK]

This is how I worked.

As you said
we know that L₁ = L₂ = L_n
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 t_n = t₁.[√n - √(n-1)]

But this way is very long i guess and I have no idea how to solve it faster.
Thank You for your help.

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]

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.

 

[HAF]

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.

Can you please give me one more hint please?

 

[PeroK]

Can you please give me one more hint please?

Does this equation help?

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

 

[HAF]

Does this equation help?

What I did this time is this.
v_n = v_n-1 .at_n
which means a=(v_n -v_n-1) / t_n

and that's what I put in equation s=v_n-1.t_n + 1/2.a.t_n²

what I got is this t_n = 2s/(v_n + v_n-1)

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

 

[PeroK]

What I did this time is this.
v_n = v_n-1 .at_n
which means a=(v_n -v_n-1) / t_n

and that's what I put in equation s=v_n-1.t_n + 1/2.a.t_n²

what I got is this t_n = 2s/(v_n + v_n-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.

 

[HAF]

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.

And what is your advise?

 

[PeroK]

And what is your advise?

$s = \frac12 at^2$

 

[HAF]

$s = \frac12 at^2$

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]

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.

 

[HAF]

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?

 

[HAF]

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]

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.

Answer:
After one second, you are dead.


Sorry. Couldn't resist. Carry on.