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Need to compute distance during during acceleration of a train

  1. Aug 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A train is traveling down a straight track at 20 m/s when the engineer applies the brakes, resulting in an acceleration of -1.0 m/s2 as long as the train is in motion. How far does the train move in a 40-s time interval starting at the instant the brakes are applied?

    2. Relevant equations

    [itex]\Delta[/itex]x = v0t + (1/2)at2

    3. The attempt at a solution

    Answer from the back of the book: 200 m, but my answer does not agree!

    [itex]\Delta[/itex]x = v0t + (1/2)at2

    [itex]\Delta[/itex]x = (20 m/s)(40 s) + (1/2)(-1.0 m/s2)(40 s)2

    [itex]\Delta[/itex]x = 800 m + (1/2)(-1,600 m)

    [itex]\Delta[/itex]x = 800 m - 800 m

    [itex]\Delta[/itex]x = 0 m


    This has to be wrong! I have gone over the calculations and the equation numerous times, and I can't find anything wrong though. Can anyone help?

    Thanks!
     
  2. jcsd
  3. Aug 30, 2013 #2

    Doc Al

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    Staff: Mentor

    Note that the problem specifies "as long as the train is in motion". When does the train come to rest?
     
  4. Aug 30, 2013 #3

    Zondrina

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    Homework Helper

    The answer at the back of your book doesn't make sense at all. How can distance be measured in ##m/s^2##?

    EDIT : I see you change it.
     
  5. Aug 30, 2013 #4
    Ok, thanks for giving me a hint Doc Al. Maybe the train came to a stop in less than 40 s, and they put in "40 s" just to trick us?

    This is what is know:

    v0 = 20 m/s
    v = 0 m/s, assuming the train did stop
    a = -1.0 m/s2
    t < 40 s, assuming the train stopped before 40 s

    If I can compute t from v0 to v, I then can compute [itex]\Delta[/itex]x.

    These are the five formulae I have to choose from (all are for constant acceleration):

    [itex]\bar{v}[/itex] = (1/2)(v0 + v)

    [itex]\Delta[/itex]x = (1/2)(v0 + v)(t)

    v = v0 + at

    [itex]\Delta[/itex]x = v0t + (1/2)at2

    v2 = v02 + 2a[itex]\Delta[/itex]x

    Unfortunately, they don't work with the info that I have.
     
  6. Aug 30, 2013 #5

    Doc Al

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    Staff: Mentor

    Use this one to figure out how long it takes for the train to come to rest.
     
  7. Aug 30, 2013 #6
    Wow, you post fast!

    Ok:

    v = v0 + at

    t = (1/a)(v - v0)

    t = (1/-1.0 m/s2)(0 m/s - 20 m/s)

    t = (-1.0 s2/m)(-20 m/s)

    t = 20 s

    I can see a problem though:

    I do not know for a fact that v (or vf) is 0 m/s, because it may have reached that speed after the 40 s. But maybe that formula finds the time from vo to v, and then you have to use that time in my original formula:

    [itex]\Delta[/itex]x = v0t + (1/2)at2

    Was my mistake assuming the time to stop the train was 40 s?

    Also, is there a way to do "strike through" formatting here? I wanted to strike though text I edited.
     
    Last edited: Aug 30, 2013
  8. Aug 30, 2013 #7

    Doc Al

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    Staff: Mentor

    You messed up the units. That should come to be seconds.

    Try it and see!

    You messed up the units, which I'm glad that you realize. Don't blame the notation!

    1/a = 1/(-1 m/s^2) = - 1 s^2/m.

    You can write fractions using Latex:
    [tex]\frac{1}{a}=\frac{1}{-1 m/s^2} = -1 s^2/m[/tex]
     
  9. Aug 30, 2013 #8
    Ok, thank you! I am going to try writing fractions with Latex later on! And I found the "strike through" button above.

    Let me try finishing the problem:

    [itex]\Delta[/itex]x = v0t + (1/2)at2

    [itex]\Delta[/itex]x = (20 m/s)(20 s) + (1/2)(-1.0 m/s2)(20 s)2

    [itex]\Delta[/itex]x = 400 m - 200 m

    [itex]\Delta[/itex]x = 200 m

    Oh my gosh! The answer matched the one in the book!

    Thank you.

    I guess my mistake was in assuming the train took the 40 s mentioned to come to a complete stop, even though all the variable did fit into the equation in my first post, and appeared to give me an answer.
     
    Last edited: Aug 30, 2013
  10. Aug 30, 2013 #9

    Doc Al

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    Staff: Mentor

    OK, much better.

    But you just found that the train came to rest in 20 seconds!

    Sure.

    Yes. Something of a trick question.

    Sure.
     
  11. Aug 30, 2013 #10
    Thank you so much Doc Al! I am learning so much through the math and physics homework help forums here!
     
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