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Math Help - A mathematical adventure down the river...!

  1. #1
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    A mathematical adventure down the river...!

    Q Tom was floating down the river on a raft, when 1 km lower down, Michael took to the water in a rowing boat. Michael rowed downstream at his fastest pace. Then he turned around and rowed back, arriving at his starting point just as Tom drifted by. If Michael's rowing speed in still water is ten times the speed of the current in the river, what distance had Michael covered before he turned his boat around?

    Okay... First I drew a diagram, and defined some variables...

    V_{M} = Michael's rowing velocity in still water

    V_{C} = Velocity of the current in the river

    T_{1} = Time taken for Michael to cover x km downstream

    T_{2} = Time taken for Michael to cover x km upstream

    V_{T} = Time taken for Tom to cover 1 km

    We know V_{M} = 10V_{C}. We also know that  S = \frac{D}{T}

    Also, the velocity of the current will affect Michael's rowing velocity. So, Michael's speed downstream will be V_{M} + V_{C} = 11V_{C}, and V_{M} - V_{C} = 9V_{C} when he turns around and moves upstream.

    So, I got three equations, as follows:

     11V_{C} = \frac{x}{T_{1}}...[1]

     9V_{C} = \frac{x}{T_{2}}...[2]

    For the third equation, I used the fact that T_{1} + T_{2} equals the time taken for Tom to cover 1 km. So:

    T_{1} + T_{2} = \frac{1}{V_{T}}...[3]

    \Rightarrow\frac{x}{11V_{C}} + \frac{x}{9V_{C}} = \frac{1}{V_{T}}

    \Rightarrow\frac{9V_{C}x + 11V_{C}x}{99V_{C}^{2}} = \frac{1}{V_{T}}

    There are too many variables... How do I proceed? Or am I on the wrong track... and there's an easier way to get this done?

    Help!

    Thanks.

    ILoveMaths07.
    Last edited by ILoveMaths07; September 6th 2008 at 10:10 AM.
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    let v = current speed
     10v = rowing speed
     t_1 = time downstream for the rower
    t_2 = time back upstream for the rower

    for the "floater" ...

     v(t_1+t_2) = 1 \, km

    for the rower ...

    11v(t_1) - 9v(t_2) = 0 \, km

    from the rower's equation ...

    \frac{11}{9}t_1 = t_2

    sub into the floater's equation ...

    v\left(t_1 + \frac{11}{9}t_1\right) = 1

    vt_1 = \frac{9}{20}

    rower's downstream distance = 11vt_1 = \frac{99}{20} \, km
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