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Another lengthy mechanics questioN!

  1. Mar 2, 2005 #1
    Taken from a physics magazine..

    A Child in a boat needs to cross the river. The speed of the current in the river is k times greater than the speed of the boat in still water. If a child crosses the river in such a way as to minimize the lateral displacement it takes time T to cross. What is the minimun time required to cross the river.

    So far with use of vector diagrams i came up with the pythag arrangement to find Tm(min)

    Just an idea :)

    [edit]tested and wrong :<[/edit]

    Open for suggestions
    Last edited: Mar 3, 2005
  2. jcsd
  3. Mar 3, 2005 #2
    I'm not sure I understand the question fully but here are my thoughts:

    - It depends on whether [tex] k < 1 [/tex] or [tex] k >= 1 [/tex].
    - Since you say it's k times greater, i'll assume k > 1 but this creates a less interesting problem than the other case.

    You can reason without the help of any formula that, if the speed of the river (which is the component of the velocity downstream) is greater than the still water speed of the boat (which is the maximum component of the velocity perpendicular to the flow of the river), there is no advantage to be gained in reducing the lateral distance by any strategy other than aiming the boat directly at the other bank. This strategy will alse lead to the minimal time to cross in the absence of other constraints (think about velocity components).

    We are given that the time for this displacement minimising strategy is T, which in this case is the same as the time minimising strategy. So it's T. Unless i've misunderstood.

    Shouldn't be a huge stretch to do this for a more general case if that's what is needed
    1. form a general velocity vector with components V and kV, where V is the still water speed of the boat. (Note: V will be at an angle [tex] \theta [/tex] to the vertical)
    2. find [tex] \theta [/tex] which minimises the displacement. The time which gives rise to this is T.
    3. as mentioned earlier, the minimum time to cross is found by driving the boat straight at the opposite bank ([tex] \theta [/tex] = 0). So it'll be S/V where V is the still water speed of the boat and which you have an expression for in terms of T from part 2.

    I think.......
    Last edited: Mar 3, 2005
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