Distance and time between 2 train stations

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The discussion revolves around calculating the distance and time for a light-rail train traveling between two stations, involving acceleration, constant speed, and deceleration phases. The train accelerates at 2.27 m/s² for 15.7 seconds, travels 1346 meters at constant speed, and then decelerates at 2.2 m/s² until it stops. Participants share equations and methods for determining the final distance and time, with one user initially miscalculating the distance. After some guidance, the user corrects their approach and successfully finds the right answers. The conversation highlights the importance of accurate calculations in physics problems.
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Homework Statement


A light-rail train going from one station to the next on a straight section of track accelerates from rest at 2.27 m/s2 for 15.7 s/ It then proceeds at constant speed for 1346 m before slowing down at 2.2 m/s2 until it stops at the station. What is the distance between the stations? and How much time does it take the train to go between the stations?

Homework Equations


I figured if i used A=v/t i could get v which i could plug into V=x/t and then get X being the distance in the start and then the distance in the middle would just be 1346 since it was given . having trouble figuring out how to get the last one

and for the time
i would think to use Δt=Δv/a

The Attempt at a Solution


ok so i got V= 35.639, and X to equal 559.5323
would the formula to find the stopping distance be Vf^2-Vi^2/2*(-2.2)?
 
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sktgurl930 said:

Homework Statement


A light-rail train going from one station to the next on a straight section of track accelerates from rest at 2.27 m/s2 for 15.7 s/ It then proceeds at constant speed for 1346 m before slowing down at 2.2 m/s2 until it stops at the station. What is the distance between the stations? and How much time does it take the train to go between the stations?

Homework Equations


I figured if i used A=v/t i could get v which i could plug into V=x/t and then get X being the distance in the start and then the distance in the middle would just be 1346 since it was given . having trouble figuring out how to get the last one

and for the time
i would think to use Δt=Δv/a

The Attempt at a Solution


ok so i got V= 35.639, and X to equal 559.5323
would the formula to find the stopping distance be Vf^2-Vi^2/2*(-2.2)?

Yes. Looks like you are on the right track.

I think your first distance is off though by a factor of 2.
 
for my distance i get 2194.2001 m
and they said it was wrong so where do you think i went wrong??
 
sktgurl930 said:
for my distance i get 2194.2001 m
and they said it was wrong so where do you think i went wrong??

I think your first distance is off though by a factor of 2. Did you catch my quick edit?
 
lol no but thanks for helping me out ur really good at this
i got it now


Thank u soo much :smile:
 
sktgurl930 said:
lol no but thanks for helping me out ur really good at this
i got it now


Thank u soo much :smile:

Happy to help.
 
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