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Minimum time for rowing across a river and running

  1. Sep 7, 2016 #1
    1. The problem statement, all variables and given/known data
    In hot pursuit of a suspect, Agent Logan of the FBI must get directly across
    a 1200m wide river in minimum time. The river's current is 0.80m/s, she can row
    a boat at 1.60m/s and she can run 3.00m/s. Describe the path she should take
    (rowing + running along the shore) for the crossing to take the minimum amount
    of time and determine the minimum time

    2. Relevant equations
    Relative velocity and SUVAT equations

    3. The attempt at a solution

    Let's make it so that the river goes horizontally from right to left, and the angle that the boat takes with the positive x-axis going across is ##\theta##.

    Then it's position vector across the river is ##\vec{r} = (v_{row} \cos \theta - v_{riv})t \hat{i} + (v_{row} \sin \theta)t \hat{j}##.

    Let ##d = 1200 ~m##.
    Then ##d = v_{row} \sin \theta t_1##
    and
    ##\displaystyle t_1 = \frac{d}{v_{row} \sin \theta}##

    Also, the change in the x-direction in the water is ##\delta x = (v_{row} \cos \theta - v_{riv})t_1##
    and the change in the x-direction on land is ##\delta x = v_{run} t_2##

    Next, I let ##\displaystyle T = t_1 + t_2 = \frac{d}{v_{row} \sin \theta} + \frac{\delta x}{v_{run}}##

    then ##\displaystyle T = \frac{d}{v_{row} \sin \theta} + \frac{(v_{row} \cos \theta - v_{riv})d}{v_{row} v_{run} \sin \theta}##

    Finally, I take ##\displaystyle \frac{dT}{d \theta} = 0##, to get the minimum ##T##, and find that it is at a minimum when ##\displaystyle \cos \theta = \frac{v_{row}}{v_{run} - v_{riv}}##.

    When I plug in the velocities, I find that ##\theta = 114.9## degrees. This is not the right answer, so what am I doing wrong?
     
  2. jcsd
  3. Sep 7, 2016 #2

    PeroK

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    And what precisely is wrong with 114.9 degrees?
     
  4. Sep 7, 2016 #3
    I look at the book, and it says that the correct answer is 24.9 degrees...
     
  5. Sep 7, 2016 #4

    PeroK

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    Hmm 114.9 degrees or 24.9 degrees? I wonder ...
     
  6. Sep 7, 2016 #5
    But what would justify me subtracting 90 degrees? Also, even if I plug 24.9 back into my equations, I am not getting the correct numbers for the other parts of the problem, which is that the minimum time is 862 seconds, and that the person runs along the bank for 104 meters...
     
  7. Sep 7, 2016 #6

    Mark44

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    Yeah, I do, too. Could there be any relationship between those two numbers?
     
  8. Sep 7, 2016 #7

    PeroK

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    I haven't studied the detail of your solution, but 114.9 (measured from parallel with the river bank) is the same angle as 24.9 (measured from a line at right angles across the river). I think you've measured the angle from parallel to the riverbank; or, at least, that's the angle your calculations assume.
     
  9. Sep 7, 2016 #8
    Well then I guess that's fine, I'll take it that somehow I am measuring parallel from the bank, and that the solution entails that I measure from the right angle to the bank. However, this still doesn't explain why inserting either 24.9 or 114.9 into the original equations does not result in me computing the minimum time, 862 seconds. For example, ##t_1 + t_2 = 635.621##, which is too small.
     
  10. Sep 7, 2016 #9

    PeroK

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    I get ##t_1 = 827s##, ##x =104m##, ##t_2 = 34.8s##

    I suspect you just made a mistake in the calculations.
     
  11. Sep 7, 2016 #10
    Could you please show me the expression you use to get ##t_1##? For the life of me I keep getting the wrong answer even though I am using the right angle...
     
  12. Sep 7, 2016 #11

    PeroK

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    I have ##t_1 = \frac{d}{vcos\theta}##

    But, that's with ##\theta = 24.9##, upstream, as measured from the line across the river.

    My ##cos\theta## should be the same as your ##sin\theta##
     
  13. Sep 7, 2016 #12
    Finally, it all checks out now. Thanks for the help!
     
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