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Thread: Distance Word Problem

  1. #1
    Super Member harpazo's Avatar
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    Distance Word Problem

    A man "A" sets out from a certain point and traveled at the rate of 6 kph. After "A" had gone two hours , another man "B" set out to overtake "A" and went 4 km the first hour, 5 km the second hour, 6 km the third hour and so gaining 1 km every hour. How many hours after "B" left were they together?

    I know the formula D = rt is needed.

    Man A:

    rate = 6 kmh

    time = x + 2

    Distance:

    6(x + 2)

    I am having trouble setting up Man B.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Distance Word Problem

    Quote Originally Posted by harpazo View Post
    A man "A" sets out from a certain point and traveled at the rate of 6 kph. After "A" had gone two hours , another man "B" set out to overtake "A" and went 4 km the first hour, 5 km the second hour, 6 km the third hour and so gaining 1 km every hour. How many hours after "B" left were they together?

    I know the formula D = rt is needed.

    Man A:

    rate = 6 kmh

    time = x + 2

    Distance:

    6(x + 2)

    I am having trouble setting up Man B.
    Note: The problem doesn't say where Man B starts (tsk! tsk!) so I'll assume both Men A and B start from the same position.

    There's probably a more "elegant" solution but we'll just have at it.
    Check out where the men are every hour.

    For instance, at the end of hour 3 Man A has traveled (6 kph)(3 h) = 18 km. At the end of hour 3 Man A has only traveled for 1 hour so he traveled (4 kph)(1 h) = 4 km.

    At the end of hour 4 Man A has traveled 18 km + (6 kph)(1 h) = 18 km + 6 km = 24 km. Man B has traveled 4 km + (5 kph)(1 h) = 4 km + 5 km = 9 km.

    Still not at the same distance. So do hour 5, 6, etc. until B catches up with A.

    By the way, if you are doing anything involving units I will get on your case if you don't use them correctly. I'm a Physicist... unit analysis is in my DNA.

    -Dan
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    Re: Distance Word Problem

    Just out of curiosity, does the DNA you are born with determine whether or not you will become a physicist or does your DNA change when you become a physicist?
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: Distance Word Problem

    Quote Originally Posted by HallsofIvy View Post
    Just out of curiosity, does the DNA you are born with determine whether or not you will become a physicist or does your DNA change when you become a physicist?
    It's destiny. Our glorious Physics gifts were given to us by the gods as existence dawned to bless the World with our power and knowledge.

    Of course standing in the beam of a particle accelerator might have something to do with it, too.

    -Dan
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    Re: Distance Word Problem

    Quote Originally Posted by harpazo View Post
    A man "A" sets out from a certain point and traveled at the rate of 6 kph.
    After "A" had gone two hours , another man "B" set out to overtake "A"
    and went 4 km the first hour, 5 km the second hour, 6 km the third hour
    and so gaining 1 km every hour.
    How many hours after "B" left were they together?
    Too busy or tired to draw a "speed = distance/time diagram" ?

    Are you expected to solve by trial/error method (as Dan suggests)
    or are you expected to solve using formula(s)?
    If by formula, you'll need the arithmetic series formula.
    The distance will be 12 + 6n representing A's distance.
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    Re: Distance Word Problem

    Quote Originally Posted by topsquark View Post
    Of course standing in the beam of a particle accelerator might have something to do with it, too.

    -Dan
    https://en.wikipedia.org/wiki/Anatoli_Bugorski
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  7. #7
    Forum Admin topsquark's Avatar
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    Re: Distance Word Problem

    Speaking seriously, I didn't know anyone could survive a proton beam. Amazing.

    -Dan
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  8. #8
    Super Member harpazo's Avatar
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    Re: Distance Word Problem

    Thank you everyone.
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  9. #9
    Super Member harpazo's Avatar
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    Re: Distance Word Problem

    Quote Originally Posted by DenisB View Post
    Too busy or tired to draw a "speed = distance/time diagram" ?

    Are you expected to solve by trial/error method (as Dan suggests)
    or are you expected to solve using formula(s)?
    If by formula, you'll need the arithmetic series formula.
    The distance will be 12 + 6n representing A's distance.
    Expected? I'm not in a math class with teacher expectations. I simply solve problems to increase my math skills, particularly with respect to word problems. Can you set up the equation(s) for me? I will do and show my work.
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  10. #10
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    Re: Distance Word Problem

    Arithmetic series formula:
    SUM = n[2a + d(n - 1)] / 2
    n = number of terms
    a = 1st term
    d = common difference between terms

    With your problem: n = ?, a = 4, d = 1
    and the SUM equals A's distance: 6n + 12 (6 miles per n + 12 miles lead).

    So:
    n[2a + d(n - 1)] / 2 = 6n + 12
    Substitute a=4 and d=1, then solve for n.
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  11. #11
    Super Member harpazo's Avatar
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    Re: Distance Word Problem

    Good notes.
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