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Velocity Vectors and Navigation Across A River

  1. Apr 16, 2017 #1
    1. The problem statement, all variables and given/known data
    Sandra needs to deliver 20 cases of celery to the farmers' market directly east across the river, which is 32 meters wide. Her boat can move at 2.5 km/h in still water. The river has a current of 1.2 km/h flowing downstream, which happens to be moving in a southerly direction.

    a) Where will Sandra end up if she aims her boat directly across the river?

    b) How far will she have to walk to reach the market?

    c) How could Sandra end up at her destination without walking?

    d) Which route will result in the shortest time for Sandra to reach her destination? Sandra can walk at 0.72 m/s when she is pulling her wagon loaded with all 20 cases of celery. She has her wagon pre-loaded on the boat.

    2. Relevant equations


    For a) ## tan\Theta = {\frac {opposite} {adjacent}}##


    b) nil


    c) ## sin\alpha = {\frac {opposite} {hypotenuse}}##


    d) ## tan\alpha = {\frac {opposite} {adjacent}}##


    ## \Delta t = {\frac {\Delta d} {\vec v}}##


    3. The attempt at a solution

    a)

    ## tan\Theta = {\frac {opposite} {adjacent}}
    \\ = {\frac {1.2km} {2.5km}}
    \\ \Theta = \tan^{-1} \left ( {\frac {1.2km} {2.5km}} \right)
    \\ = 25.6##

    26 degrees

    Next, the distance.

    ##tan\Theta = {\frac {opposite} {adjacent}}
    \\tan26 = {\frac {\vec d_2} {32m}}
    \\ \vec d_2 = 15.6##

    Sandra ends up 16 meters [East 26 degrees South] on the opposite shore.

    b) She has to walk 16 meters north to get to the market

    c) She needs to aim her boat to the Northeast.

    ##sin\alpha = {\frac {opposite} {hypotenuse}}
    \\ \alpha = \sin^{-1} \left ( {\frac {1.2km} {2.5km}} \right)
    \\ \alpha = 28.7##

    She needs to aim East 29 degrees North

    d) ##tan\alpha = {\frac {opposite} {adjacent}}
    \\ adjacent = {\frac {opposite} {tan\alpha}}
    \\ = {\frac {1.2km} {tan29}}
    \\ = 2.16 km/h##

    conversion to m/s

    ## {\frac {(2.16)(1000)} {3600}}
    \\ = 0.6 m/s##


    Finally, comparison of sailing vs walking speeds

    ## \Delta t = {\frac {\Delta d} {\vec v}}
    \\ = {\frac {32m} {0.6m/s}}
    \\ = 53.3##

    ## {\frac {16m} {0.72m/s}}
    \\ = 22.2##

    It takes her 22 seconds to walk from 16m away versus the 53 seconds it would take her to sail directly there.

    Am I correct in any of this?
     
  2. jcsd
  3. Apr 17, 2017 #2

    scottdave

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    All except in the last part, you need to add to her walking time, the time it took to cross the river.
     
  4. Apr 17, 2017 #3
    I realized that I left out sailing time to add to the walking time about two seconds after I hit post. Thank you for confirming my work.
     
  5. Apr 17, 2017 #4

    scottdave

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    Also, on part A, you don't really have to take the inverse tangent, then the tangent, again. If you realize that tangent function returns the slope of a line with that angle. The "slope" of the path ( a proportion). You travel 1.2 km South, for every 2.5 km East. So it is (1.2 km) / (2.5 km) = 0.48 [dimensionless] Now we just need to multiply this by the distance (East) across the river, and you will know how far South you moved.
     
  6. Apr 18, 2017 #5
    That's how the textbook taught to find the distance: use trigonometry to get the distance and inverse to get the angle. I didn't know your method, but thank you for teaching me that. I don't think I need the inverse angle, but I decided to throw it in my work as the single solitary example problem in the text included it.
     
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