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Math Help - man on bridge

  1. #1
    mathionyness
    Guest

    Exclamation math HW help

    Here's the problem:
    A man is 2/5 the way walking across a railroad bridge when he hears a train approaching from behind him. He can reach either end of the bridge just as the train reaches the same point. If the train is traveling 60 mph, at what average speed must the man run?
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    This is a nice question.

    Draw the figure on paper.
    It is a horizontal line segment ABCD. The train is at point A; the start or near end of the bridge is at point B; the man is at point C; and the end or far end of the bridge is at point D. We don't know the lengths of AB (distance of train from the near end of the bridge) and BD (the bridge), so let AB = X miles and BD = L miles long.
    BC is 2/5 of L, or 2L/5 miles, and so CD is 3/5 of L, or 3L/5 miles.

    The man has two choices: proceed and run for the far end, or go back and run for the near end. In both choices, he and the train would reach the same end at the same time. [Meaning, the hapless man would die at any of the two ends! So why run? Why wouldn't he just stay where he is, sing a song, and die there. :-)]

    Choice #1. He goes back to the near end.
    Time spent = t1 hours

    distance = rate * time

    train, x = 60*t1 ---(1)
    man, 2L/5 = v*t1 ---(2), where v = average speed of man's running.

    Choice #2. He goes ahead and run for the far end.
    Time spent = t2 hours

    distance = rate * time

    train, x+L = 60*t2 ----(3)
    man, 3L/5 = v*t2 -----(4)

    That is it. Only 4 independent equations but there are 5 unknowns ---v, x, L, t1, and t2.
    Interesting. Supposed to be there is no unique solution but I found it.

    We are looking for v, so we try to eliminate x,L,t1,t2.

    We can eliminate t1 from (1) and (2).
    From (1), t1 = x/60
    From (2), t1 = 2L/5v
    t1 = t1,
    x/60 = 2L/5v
    Cross multiply,
    x*5v = 60*2L
    vx = 24L ------(5)

    Likewise, we can eliminate t2 from (3) and (4).
    From (3), t2 = (x+L)/60
    From (4), t2 = 3L/5v
    t2 = t2,
    (x+L)/60 = 3L/5v
    Cross multiply,
    (x+L)*5v = 60*3L
    (x+L)v = 36L -----(6)

    We can eliminate L from (5) and (6).
    From (5), L = vx/24 ---***

    From (6),
    vx +vL = 36L
    vL -36L = -vx
    L(v-36) = -vx
    L = -vx/(v-36)
    L = vx/(36-v) ---***

    L = L,
    vx/24 = vx/(36-v)
    Divide both sides by vx,
    1/24 = 1/(36-v)
    24 = 36 -v
    v = 36 -24
    v = 12 mph --------answer.

    -----------
    The idea was for the man to beat the train at any of the two ends of the bridge. Apparently, the problem means if the man and the train reach the same end of the bridge at the same time, the man will be saved.
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  3. #3
    Newbie
    Joined
    Sep 2005
    Posts
    6

    hmmmm

    on steps 5 and 6 you can make it easier by isolating v so that you have the 2 equations

    v=36L/(x+L)
    v=24L/x

    that way you get 36L/(x+L)=24L/x

    cross multiply and solve for x

    you get x/(x+L)=2/3 => 2x+2L=3x => x=2L

    how you throw that x on either of the two velocity equations and you get

    v=24L/2L => v=12
    or
    v=36L/(L+2L) => v=36L/3L => v=12

    I only like this way better cuz then you dont have to worry about negative variables and for some reason it looks simpler to me... but hail for ticbol for solving it first
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