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    lalav84's Avatar
    lalav84 Posts: 2, Reputation: 1
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    #1

    Jan 24, 2007, 01:08 PM
    Word problem
    Two trains are headed right for each other they are 180 miles apart. Train A is going 50mph, train B is going 40 mph train A starts at point A. when will they collide and where will they collide in reference to point A?

    I don't even know where to start on this problem, it just is confusing me... please help!! Thnaks
    JChev06's Avatar
    JChev06 Posts: 9, Reputation: 1
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    #2

    Jan 24, 2007, 02:16 PM
    They will collide in 2 hours, they will be 100 miles from point A.

    Just multiply each trains speed by 2 (that equals 180 miles) and that's how you get the answer.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #3

    Jan 24, 2007, 02:16 PM
    write an equation for x (distance apart)

    x = 180 - (50+40)t

    this gives you the distance they are apart, solve for x = 0 to find how many hours (t) it takes for them to collide.

    Then you can use speed = distance/time to find the distance train A has travelled from point A until they collide (at time t)

    If you need further help, please ask.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #4

    Jan 24, 2007, 02:19 PM
    JChev06, although your answer is correct, please read This Announcement for the site's policy on helping people with homework
    JChev06's Avatar
    JChev06 Posts: 9, Reputation: 1
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    #5

    Jan 24, 2007, 02:31 PM
    Sorry, I was not aware of that. Won't happen again.
    CaptainForest's Avatar
    CaptainForest Posts: 3,645, Reputation: 393
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    #6

    Jan 24, 2007, 02:57 PM
    JChev06,

    That is not an official policy that you have to abide by.

    I don't always.

    If you wish to help people with their homework, that is your decision.

    Sometimes I do, but other times, when they don't even post how they attempted it, I don't bother to.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #7

    Jan 24, 2007, 11:30 PM
    Sure, you don't need to bother, but if you do, stating how to get the answer is much more useful than the answer itself
    newb's Avatar
    newb Posts: 14, Reputation: 1
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    #8

    Jan 25, 2007, 06:32 AM
    OK, I'm going to show you a way to do it that will make all of these problems look the same. You're teacher will probably make a chart, unfortunately, to do this problem. My method is similar to Capuchin's method. You need to know to do two things to solve these types of problems: 1) convert what they give you into a statement about lengths of line segments; and 2) know that distance = rate times time (d = rt). That's it! Let's do 1) first.

    Da Db
    1) A_______________x__________B

    <---------------180---------------->

    The picture show that line segment 180 = Da + Db (Where Da is distance that a travels before the collision and Db is the distance that b travels before the collision).

    Now replace Da with RaTa and Db with RbTb. So the above equation becomes 180 = RaTa + RbTb. Now plug in what you know, namely Ra = 50 and Rb = 40. The equation becomes 180 = 50Ta + 40Tb. Notice that train A leaves point A at the same time that train B leaves point B. The key point is that when they collide they've been on the tracks the same amount of time since they left their respective points A and B. So we can write Ta = Tb = T and the equation becomes 180 = 50T + 40T. Now solve for T. Notice that if you delete all the words, it takes about 4 lines to solve the problem!

    Phil
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #9

    Jan 25, 2007, 06:48 AM
    Hi newb, thanks for explaining how I got my equation :)

    Why do you use "rate" instead of "speed"?
    Speed is more specific, as rate can be interpreted as many many things.
    newb's Avatar
    newb Posts: 14, Reputation: 1
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    #10

    Jan 25, 2007, 08:21 AM
    Hi Capuchin,

    This IS a rate. That is to say it has both magnitude (speed) and most certainly direction. That's why we refer to these as Distance/Rate/Time problems instead of Distance/Speed/Time problems and we write d = rt, not d = st.

    Phil
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    newb Posts: 14, Reputation: 1
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    #11

    Jan 25, 2007, 08:25 AM
    I forgot to say that rate is more specific when we do these problems since it implies magnitude and direction instead of just magnitude (speed).
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #12

    Jan 25, 2007, 08:41 AM
    The vector for speed is velocity, the term rate is utterly ambiguous.

    I have never seen anyone use rate meaning velocity ever before. And I have never seen anyone use r in an equation for velocity, as this would be confused with radius.

    Rate does kind of make sense, but not as a vector like you are claiming. For example, what direction is a heart rate? This is clearly an example of rate measuring speed, and is also clearly a scalar.

    As far as I can see, a rate is a measure of anything over a length of time. Meters per second, beats per minute, Joules per second could all be called rates. Using it in this context is not more specific than speed, as speed is used solely for measuring a change in position over time. Furthermore it is a scalar. Velocity is what you should have used for the vector.
    newb's Avatar
    newb Posts: 14, Reputation: 1
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    #13

    Jan 25, 2007, 09:09 AM
    Capuchin, I'm very surprised that you have never seen this before. Have a look at the following: Math Forum: Ask Dr. Math FAQ: Distance, Rate, and Time. I understand your claim regarding it being ambiguous, however as I said, within the confines of this problem, it most certainly is not. I agree that velocity is more specific outside the context of this problem though and is used more commonly in classical mechanics. However, I emphasize rate precisely to show that this notion can be generalized to other types of (rate not velocity or speed) problems where you replace the d with something else. In truth I have never heard these problems referred to as anthing other than distance/rate/time problems.


    Phil

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