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

In summary: 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.
  • #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.
 
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  • #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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