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

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SUMMARY

The discussion revolves around calculating the time it takes for the n-th wagon of a train, which accelerates uniformly, to pass a stationary observer. The first wagon passes in 4 seconds, and it is established that the acceleration is constant and all wagons have the same length. The derived formula for the time taken by the n-th wagon is tn = t1 * (√n - √(n-1)), where t1 is the time for the first wagon. The conversation emphasizes the importance of understanding kinematic equations and the assumptions made regarding the problem's parameters.

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  • Understanding of kinematic equations, specifically s = v0.t + 1/2at² and v = v0 + at.
  • Knowledge of constant acceleration concepts in physics.
  • Familiarity with algebraic manipulation to derive formulas.
  • Basic understanding of sequences and their application in problem-solving.
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  • Learn how to apply algebraic techniques to solve physics problems involving sequences.
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  • #31
HAF said:
Can you please give me one more hint please?

Does this equation help?

HAF said:
s= vo.t + 1/2at^2
 
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  • #32
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.
 
  • #33
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.
 
  • #34
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.
And what is your advise?
 
  • #35
HAF said:
And what is your advise?

##s = \frac12 at^2##
 
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  • #36
PeroK said:
##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.
 
  • #37
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.
 
  • #38
mfb said:
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?
 
  • #39
mfb said:
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!
 
  • #40
HAF said:
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.
:biggrin:

Sorry. Couldn't resist. Carry on.
 
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