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Alternate River crossing problem

  1. Sep 14, 2010 #1
    Hello, we are a few students that have been strugling with this question for some time now. We see no other solution than to ask you for help. Some clues would atleast help us on the way:) Here follows the question:

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

    A man wants to cross a river that is 500 meters wide. He can row with a speed of 3.0km/h relative to the water. The river flows with a speed of 2km/h, while the mans walkingspeed at land is 5km/h

    1.)What is the optimal angle to cross the river at with regards to minimizing the time used.
    2.)How long did it take?

    Thanks in advance
     
  2. jcsd
  3. Sep 14, 2010 #2
    Lets see, looking at the clues (walking speed in the land) I think that you have to compare between two options:
    a) Row in straight line, shorter but slower. You decompose V in Vx and Vy, with the sen of the angle. (42º; Vx = crossing speed 2,2 km/h, Vy = 2 km/h enough to row straight ahead)
    b) The other option its to begin to tow with an angle of 0º; he will go faster, but will arrive far away, having to walk to the beginning point. Vx=3, time to arrive = (500m) 10 minutes.
    But in ten minutes, at 2km/h he will move river down 320 meters, that will be 3.84 min walking river up. Total = 13.84 minutes.
    In option a) 500 m at 2.2 km/h 13.63 minutes
    And maybe you can optimize the result derivating, but I am hurry now, sorry! ^.^
     
  4. Sep 14, 2010 #3
    We have been trying all those options before. If i reveal that the answer should be 25,4 degrees and 12,7 min. Is that any help:P?
     
  5. Sep 14, 2010 #4

    kuruman

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    Homework Helper
    Gold Member

    You need to find the crossing time t1 and the walking time t2 and minimize the sum of the two by taking the derivative with respect to θ. I did it and it gives the answer you posted for the angle. If you say you tried it and it didn't work, then post what you did and perhaps someone can find where you might have gone wrong.
     
  6. Jan 30, 2012 #5
    How do I get the expression for the two times then? When I use the forumla s=v*t with vectors, I get an expression where i have to divide the position and velocity vector?

    To find the max/min values shall not be a problem, I just can't understand how to get the expression to derivate...?
     
  7. Jan 30, 2012 #6
    What i know is the following(distance in km and speed in km/h).

    The boats speed in the x-direction:

    vx,b=3cosθ

    The boats speed in the y-direction:

    vy,b=3sinθ-2

    The distance to row:

    srow=0.500/cosθ

    The walking speed in x-direcrion:

    vx,w=0

    The walking speed in y-direction(defined positive y-direction towards the flowing river):

    vy,w=5

    Distance to walk:

    swalk=srowsinθ

    Expression for time should then be:

    ttot=trow+twalk=(srow/vrow)+(swalk/vwalk)

    My problem now is that I cannot see how to combine these equations(especially not the two speed components of the boat). Is there anyone that can give me a hint?
     
  8. Jan 31, 2012 #7
    No one?
     
  9. Jan 31, 2012 #8

    NascentOxygen

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    Staff: Mentor

    You know the x component of his rowing speed, so determine the time to travel 0.5km.

    During this time, his y-direction speed takes him a distance downstream. Determine this distance. Now, determine the time taken to walk back over this distance.

    Add the two times together. The only unknown in this expression for total time is angle θ.
     
  10. Jan 31, 2012 #9
    Ah, thank you very much! I get it now.
     
  11. Jan 31, 2012 #10
    Well, I at least found an expression for the time:

    t=[itex]\frac{0.500}{3cosθ}[/itex]+[itex]\frac{0.500(3sinθ-2)}{15cosθ}[/itex]

    By derivating the expression I then get

    [itex]\frac{dt}{dθ}[/itex]=[itex]\frac{0.500}{3}[/itex][itex]\frac{d}{dθ}[/itex]([itex]\frac{1}{cosθ}[/itex])+[itex]\frac{0.500}{15}[/itex][itex]\frac{d}{dθ}[/itex]([itex]\frac{3sinθ-2}{cosθ}[/itex])=[itex]\frac{0.500}{3}[/itex][itex]\frac{tanθ}{cosθ}[/itex]+[itex]\frac{0.500}{15cosθ}[/itex][[itex]\frac{3}{cosθ}[/itex]-2tanθ]

    To find the max/min values i should now set the derivated expression equal to zero, and solve for θ. My problems starts here, because I am not able to solve for θ. Have i done something wrong, or do I just suck in math?
     
  12. Feb 1, 2012 #11

    NascentOxygen

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    Staff: Mentor

    Before plunging into calculus, collect like terms. Remove the brackets, and there are 2 fractions here with cosθ in their denominator.
     
  13. Feb 3, 2012 #12
    Yes, I know, I managed to solve it. Thanks anyway!
     
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