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Math Help - Derive equation of runner

  1. #1
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    Derive equation of runner

    Hi,

    An athlete usually runs 80km at a steady speed of vkm per hour. He decides to reduce his speed by 2.5km per hour resulting in his run taking an extra 2 hours and 40 minutes.
    Derive \frac{80}{v}+\frac{8}{3}=\frac{160}{2v-5}

    Here's what I did which unfortunately is wrong:

    \frac{80}{60(v-2.5)}-\frac{80}{60v}=160 (Converted it to minutes)
    \frac{80}{60(v-2.5)}=160+\frac{80}{60v}
    \frac{80}{60v-150}=160+\frac{80}{60v}
    80=160(60v-150)+\frac{80(60v-150)}{60v}
    80=9600v-24000+\frac{80(60v-150)}{60v}
    24080-9600v=\frac{80(60v-150)}{60v}
    60v(24080-9600v)=80(60v-150)
    1444800v-576000v^2=4800v-12000
    576000v^2-1440000v-12000=0
    48v^2-120v-1=0

    How on Earth did I get it so far from the answer?

    Any help is appreciated
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  2. #2
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    Quote Originally Posted by webguy View Post
    Hi,

    An athlete usually runs 80km at a steady speed of vkm per hour. He decides to reduce his speed by 2.5km per hour resulting in his run taking an extra 2 hours and 40 minutes.
    Derive \frac{80}{v}+\frac{8}{3}=\frac{160}{2v-5}
    (time in hrs it normally takes) + (2 hrs + 40 min) = (longer time when he runs slower)

    \frac{80}{v} + \frac{8}{3} = \frac{80}{v-2.5}

    clear the decimal in the last fraction ...

    \frac{80}{v} + \frac{8}{3} = \frac{160}{2v-5}

    common denominator is 3v(3v-5) ...

    \frac{80 \cdot 3(2v-5)}{3v(2v-5)} + \frac{8v(2v-5)}{3v(2v-5)} = \frac{160 \cdot 3v}{3v(2v-5)}

    numerators form the equation ...

    240(2v-5) + 8v(2v-5) = 480v

    30(2v-5) + v(2v-5) = 60v

    60v - 150 + 2v^2 - 5v = 60v

    2v^2 - 5v - 150 = 0

    (2v + 15)(v - 10) = 0

    only solution which works in the context of the problem is v = 10 km/hr

    at 10 km/hr, takes 8 hrs to run 80 km

    at 7.5 km/hr, takes 10 hrs 40 min
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \frac{80}{v} + \frac{8}{3} = \frac{80}{v-2.5}

    clear the decimal in the last fraction ...

    \frac{80}{v} + \frac{8}{3} = \frac{160}{2v-5}
    How did you do that step? It looks like you only multiplied one side by \frac{2}{2}.

    Also, why is my answer so incorrect. The logic of my first step seems correct in my mind; is it not?
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  4. #4
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    Quote Originally Posted by webguy View Post
    How did you do that step? It looks like you only multiplied one side by \frac{2}{2}.

    correct ... 2/2 = 1 ... multiplication by 1 changes nothing.


    Also, why is my answer so incorrect. The logic of my first step seems correct in my mind; is it not?

    your "conversion" to minutes (which is not necessary to begin with) is incorrect. note that every term in the first equation represents time in hrs.

    conversion to minutes requires multiplying each term by 60 min/hr.

    ...
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  5. #5
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    But I converted every hour to minutes. Even though I was working in minutes I don't see how what I did was wrong. Would you mind pointing out the specific flaws of my answer?
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  6. #6
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    Quote Originally Posted by webguy View Post
    But I converted every hour to minutes. Even though I was working in minutes I don't see how what I did was wrong. Would you mind pointing out the specific flaws of my answer?
    no, you didn't ... I told you in my previous post that each term is a time in hours.

    specifically, using the first term of the original equation ...

    \displaystyle \frac{80 \, km}{v \, km/hr} = a specific amount of time in hours

    \displaystyle \frac{80 \, km}{v \, km/hr} \cdot \frac{60 \, min}{hr} = that same time above in minutes
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  7. #7
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    So it should be \frac{4800}{60v}?
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  8. #8
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    Quote Originally Posted by webguy View Post
    So it should be \frac{4800}{60v}?
    no
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  9. #9
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    Quote Originally Posted by skeeter View Post
    no
    Could you elaborate?
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  10. #10
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    Quote Originally Posted by webguy View Post
    Could you elaborate?
    Maybe someone with the time and inclination will elaborate in the fullness of time. In the meantime, you need to carefully reflect on the help you've already been given and show that you have taken on board this help.
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  11. #11
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    Quote Originally Posted by skeeter View Post
    \displaystyle \frac{80 \, km}{v \, km/hr} \cdot \frac{60 \, min}{hr} = that same time above in minutes
    60\times(\frac{80}{v}) which gives \frac{4800}{v}.

    So my original should be:

    \frac{4800}{v-2.5}-\frac{4800}{v}=160

    If this is wrong feel free to say with no explanation. I'll end this thread here at that point since I doubt I'll ever understand. Unfortunately I fail to understand why multiplying the hours by 60 minutes and dividing into the distance is incorrect, but multiplying the distance by the minutes is correct.
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