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Two Different Accelerations, find distance these trains meet

  1. Jan 31, 2010 #1
    1. Problem Statement

    Two trains face each other, adjacent tracks.
    Both at rest, and front ends are 40m apart.
    Left train accelerates at 1.0m/s^2.
    Right train accelerates at 1.3m/s^2.

    A) How far does train on the left travel before the front ends of the trains pass?

    B) If trains are each 150m in length, how long after the start are they completely past one another, assuming their accelerations are constant?



    2. Relevant equations

    x = v*t+.5*a*t^2
    vf = vi+a(t)
    x = vi*t+.5a*t^2

    A drawing:
    http://www11.picfront.org/token/y4KU/2010/02/01/1751429.jpg [Broken]




    3. Attempt to reach answer.

    A)Will this approach work? To solve , we want to get xA = xB , A and B trains on each side, and x = distance.

    so setting (vi*t+.5a*t^2)A = (vi*t+.5a*t^2)B

    What i did was use:
    x = v*t+1/2*a*t^2
    40m = 0 + .5*1.0*t^2
    Solve for t, t= sqrt(80/1) => 8.95seconds

    Next use vf = vi+a(t)
    vf = 0+a(t)
    vf = 0 + (1.0)(8.95) => 8.95m/s

    Vice versa for b train, vf = 10.20m/s , t = 7.85 seconds



    With v obtained, I have no idea what to do? A tip?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 31, 2010 #2
    Oh, I tried this as well.

    xA = 1/2*a*t^2
    xB = 1/2*a*t^2

    Solve for the time when their positions are equal.
    .5*1*t^2 = .5*1.3*t^2 . Cannot get t.

    So i tried this since I got velocity average for A

    xA = v*t , xB = 1/2*a*t^2, solve for t.

    4.47(t) = .5*1.3*t^2

    4.47/.5*1.3 = t ; t = 6.87seconds.

    Plug back into xA = v*t xA = 4.47*6.87 = 30.70m is point where train A meets train B. Logic looks good?
     
  4. Jan 31, 2010 #3
    The easiest way I see is making your picture a displacement number line. Call the far left end 0 and the far right end 40.
    Then use the train on the left and call the distance it travels when the trains pass X. Then the other train, B, would travel -(40 - X) when they meet. Just have to be careful and make sure train B's acceleration is negative. Now they will meet at the same time t so you have two equations with two variables and t is the same for both. Then its just algebra. Solve for X.

    Oops I see you have posted again. maybe this helps, maybe not.
     
    Last edited by a moderator: May 4, 2017
  5. Jan 31, 2010 #4
    Relook agained and though about displacement.

    So what I did for solving A)
    A) How far does train on the left travel before the front ends of the trains pass?

    xA + xB = 40m
    x = 1/2*a*t^2

    Let's solve for time.


    .5*(1)(t^2) +.5*1.3*t^2= 40
    t = 5.89s

    Plug back into xA = .5*1*5.89 = 17.34m <- answer ?

    Part b)
    B) If trains are each 150m in length, how long after the start are they completely past one another, assuming their accelerations are constant?


    xA = 150m and xB = 150m

    Let's find vF.
    vF = Vi+a(t)
    VF = 0 + 1(5.89) = 5.89m/s of train A.
    VF = 0+1.3(5.89) = 7.65m/s of train B


    Okay, now let's find time that train A went 150m
    x = vi*t+.5*a*t^2
    150 = 5.89(t) + .5*1*t^2
    t = 12.45sec

    and let's find time that train B went 150m
    x = vi*t+.5*a*t^2
    150 = 7.65(t) + .5*1.3*t^2
    t = 10.40seconds

    Okay, so the time they are completely past one another is at 12.45seconds or 10.40seconds? I am confused.

    I am guessing it would be 12.45sec since Train A is slow and would finally hit 150m at that time where train B already hit 150m at 10.4s
     
  6. Feb 1, 2010 #5
    part a is now correct.

    for part b:

    if train A has moved d meters at the moment the trains completely pass each other, so the position of the front of A is d meters to the right of the origin.

    where is the back of A?

    where is the back of B?

    where is the front of B?

    how much has B moved?

    what is the sum of the total movement of both trains?

    does this look like something you already solved?
     
  7. Feb 1, 2010 #6
    Hm.. If i'm understanding this correctly:

    Consider front of A the origin. Back of A is -150m from origin.
    front of B is the origin. Back of B is +150m from origin.

    The time in which I solve tells me the following:


    Okay, now let's find time that train A went 150m
    x = vi*t+.5*a*t^2
    150 = 5.89(t) + .5*1*t^2
    t = 12.45sec

    and let's find time that train B went 150m
    x = vi*t+.5*a*t^2
    150 = 7.65(t) + .5*1.3*t^2
    t = 10.40seconds

    @t = 12.45seconds, The total train has crossed the origin for A(Back of A reached front of A)
    @t = 10.40seconds, The total train has crossed the origin for B. (Back of B reached front of B)

    So @10.40 , only one train is passed the initial point(b), the end of A does not meet with the end of B yet. It is off by 2.05seconds @ that point.

    @12.45s, both trains have finally passed the initial point(a). Hm, I don't believe end of A and end of B meet up @ 12.45 since end of B has left the origin.

    I'll figure it out some more.. its getting late
     
  8. Feb 1, 2010 #7
    Beh just figured out what I did those times are when the train hits 150m. NOT what the question asked.
    Ask for When the ends of the trains are meet.

    They will meet when xA+xB=300m.

    Consider the starting point at when trains front ends face one another. some velocity on each train side.
    300 = 5.89(t) + .5*t^2 + 7.65t + .5*1.3t^2

    t = 11.30s , so at 11.30 + 5.89 = 17.19s , both ends will pass one another.
     
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