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One Dimension Kinematics Question

  1. Jun 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A fugitive tries to hop a freight train traveling at a constant speed of 6.0 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a=4.0 m/s2
    to his maximum speed of 8.0 m/s. (a) How long does it take him to catch
    up to the empty box car? (b) What is the distance traveled to reach the
    box car?



    2. Relevant equations
    v = v(initial) +at
    x = x(initial) + v(initial)t + .5at^2
    v^2 = v^2(initial) + 2a(x - x(initial)
    x=vt



    3. The attempt at a solution

    I first found out how long it took the fugitive to get to 8.0 m/s since this is his maximum speed. so i used the first equation above 8.0= 0 +4.0t... t=2s.

    now i know it takes the fugitive 2 seconds to reach max velocity. now im trying to figure out how far the fugitive traveled and how far the train traveled. For the fugitive i used the second equation above x= 0 + 0 + .5(4)(2^2) x=8m

    Now I want to find the distance the train traveled:
    x =vt
    x = 6(2) = 12

    So Now Im left with the problem that the fugitive traveled 8m in 2 seconds and the train traveled 12m in 2 seconds. a difference of 4m. So i looked at the question and said to myself since the fugitive is traveling at a higher velocity than the train the will eventually meet each other. but how to find that time??

    I broke the fugitives distance into two parts
    1. getting up to speed
    2. traveling a constant speed

    but I have no clue on what equation to use or how to approach the problem...

    p.s. sorry this is sooo long.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 10, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi iurod! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    You've found that, after 2s, he is 4m behind the train.

    From then on, he is going 2m/s faster than the train, so the time to catch it is … ? :wink:
     
  4. Jun 10, 2010 #3
    Thanks for your super fast reply Tiny Tim..


    so I came up with this but im not entirely sure if this is right

    distance of the man accelerating + velocity(t2) = Velocity of the train (t1 +t2)

    4m/s^2 + (8t) = 6(2 + t2)
    4 + 8t = 12 +6t
    2t = 8

    t = 4 seconds????

    if this is correct then the distance that the fugitive had to travel is:

    d = vt
    d = 6(4)
    d = 24 meters

    is this correct?
     
  5. Jun 10, 2010 #4

    tiny-tim

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    Sorry, I'm not following this …

    you seem to be using t to mean two different things. :confused:

    Write it out again, more carefully. :smile:
     
  6. Jun 10, 2010 #5
    yeah it doesn't make that much sense when I wrote it.

    what I meant was:

    the distance of the man accelerating + the velocity of the man(time2) = Velocity of the train (time 1 + time 2)

    I need two times because it takes the fugitive 2 seconds to reach maximum velocity, after that I set the:

    .5(acceleration)(2 seconds squared; that it took to get to maximum velocity) + the distance of the fugitive (vt) (i dont know this time so im calling it time 2) equal to the velocity(time 1 + time 2) of the train.

    so:
    .5(a)(time1)^2 + v(time2) = v(time1 + time2)

    .5(4m/s^2)(2^2) + 8m/s (time 2) = 6m/s (2+time 2)
    8 + 8time2 = 12 +6time2
    2time2 =4
    time2 = 2

    time1 + time2
    2+2 = 4 seconds


    if this is correct then the distance that the fugitive had to travel is:
    d = vt
    d = 6(4)
    d = 24 meters


    hopefully this makes sense, its hard putting my thoughts into words. hopefully its correct also :)
     
  7. Jun 11, 2010 #6

    tiny-tim

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    hi iurod! :wink:

    Yes, that's much better! :smile:

    It's a bit long-winded, though …

    you didn't need to define two "time"s, since you'd already worked out that the first one was 2 s (so it's no longer unknown, and there's no point giving names or labels to things that are known) …

    so you could have immediately called the second one "t" (instead of "time2") …

    alternatively, you could have simply said "after 2 s, the fugitive is 4 m behind, and is catching up at a constant relative speed of 2 m/s, so the further time he takes is 4/2 s = 2 s, making a total of 4 s" :wink:
     
  8. Jun 11, 2010 #7
    Re: One Dimension Kinematics Question SOLVED

    Tiny Tim,
    Your explanation makes a whole lot more sense than what I did :)

    I really appreciate all your help on this, I think PF is going to be my new home for Physics Questions :)

    Thanks again!
     
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