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Math Help - rate*time=distance

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    Talking rate*time=distance

    A passenger train and a freight train start at 405 miles apart and travel towards each other. the rate of the passenger train is twice the rate of the freight train. in 3 hours they pass each other (or collide). how fast are they going?
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    Quote Originally Posted by blame_canada100 View Post
    A passenger train and a freight train start at 405 miles apart and travel towards each other. the rate of the passenger train is twice the rate of the freight train. in 3 hours they pass each other (or collide). how fast are they going?
    Pet Peeve Warning!!! Danger Will Robinson! Danger! Danger!

    Why can't Math books just use the term "speed" like everyone else? "rate*time = distance." "rate" of what, I ask you???

    Pet Peeve over.

    Okay. Let's assume that the passenger train starts from the left and the freight train starts from the right. For the record I am going to set an origin at the starting position of the passenger train and set the +x direction to be to the right.

    The passenger train will move with a speed v_p and cover a distance x miles. The freight train will move with a speed v_f and covers a distance 405 - x miles. They meet at the same time, t = 3 hours.

    So for the passenger train:
    x = v_pt

    and for the freight train:
    405 - x = v_ft

    and
    v_p = 2v_f

    Put the first equation into the second:
    405 - (v_pt) = v_ft

    and now insert the 3rd equation into the equation we just got:
    405 - (2v_f)t = v_ft

    Now we can solve for v_f:
    405 - 2v_ft = v_ft

    405 = 3v_ft

    v_f = \frac{405}{3t} = \frac{405 \, mi}{3 \cdot 3 \, h} = 45 \, mi/h

    So v_p = 2v_f = 90 \, mi/h

    -Dan
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    Quote Originally Posted by blame_canada100 View Post
    A passenger train and a freight train start at 405 miles apart and travel towards each other. the rate of the passenger train is twice the rate of the freight train. in 3 hours they pass each other (or collide). how fast are they going?
    Let s be the speed of the frieght train.
    Then the speed of the passenger train is 2s.

    The average speed of both trains is:
    (s+2s)/2=(405/3)/2 <---- 405 / 3 = distance / time
    (3s)+2=135/2
    1.5s = 67.5

    The speed of the frieght train:
    1.5s = 67.5
    s = 67.5 / 1.5
    s = 45

    The speed of the passenger train:
    1.5s = 67.5 <---- solving for 2s
    s = 45
    2s = 90

    The speed of the freight train is 45 mph.
    The speed of the passenger train is 90 mph.
    Last edited by The Pondermatic; November 20th 2006 at 05:06 PM.
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    thx

    thx! i get it now!
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