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Vector - Need explanation

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data

    A girl can row a boat at 3 m/s in still water. She wishes to cross a 165m wide river in which the current flows at 1.5 m/s. Find the direction in which the boat must be steered in order to cross the river:

    a) by the shortest possible route

    b) in the shortest possible time

    2. Relevant equations



    3. The attempt at a solution

    The answer from my book says that for

    a) In order to cross the river by the shortest possible path, the actual path must be at right angles to the banks, which is θ=cos-1 1.5/3 = 60º

    b) in order to cross in shortest possible time, the boat must be steered at right angles to the bank, i.e θ=90º.

    But i don't really understand. Can anyone elaborate more for me...?

    Why in order to cross the river by the shortest possible path, the actual path must be at right angles to the banks?

    and why in order to cross in shortest possible time, the boat must be steered at right angles to the bank?

    Please guide me. :confused:
     
  2. jcsd
  3. Nov 1, 2011 #2

    HallsofIvy

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    Staff Emeritus
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    First, geometrically, the shortest path across a river, represented by two parallel lines, is perpendicular to the banks. If you need more detail, any other straight line across the river would form the hypotenuse of a right triangle having the perpendicular route as a leg. And [itex]c^2= a^2+b^2[/itex] so the hypotenuse is always longer than either leg. And any curved path is longer than the straight line between the two endpoints.

    But in order to actually go across the river perpendicular to the banks, you must "point" the boat upstream to overcome the downstream flow of the river. That is, the velocity vector of the boat must have two components, one upstream, equal to but opposite the speed of the river, in order to cancel it, and one across the river that actually contributes to the motion of the boat. The time it takes to cross the river, along this shortest path, is the width of the river divided by cross river component of the boat's speed.

    If, instead, you point the boat straight across the river, while letting the river take the boat downstream, with the same boat speed as above, all of the boats "still water" speed will go to that cross river component rather than just the speed minus the component required to cancel the river's speed. Of course, this assumes that you put the same effort, rowing, sailing, engine, etc. into maintaining the same "forward" speed for the boat.
     
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