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Vectors and Kayaking?

  1. Feb 13, 2017 #1
    Hello,

    Could someone please check my answer to this problem involving vectors. I have a feeling my answers are totally off; especially for B and C. All my work and applicable equations are shown in the pictures below. If my answer is wrong, please provide me with guidelines for solving the problem. I'm not expecting anyone to just give me the answer :)

    Thanks!

    Problem 8

    You are in your kayak, paddling across the Charles River, which we can treat as having parallel banks separated by a distance of 480. m. The river is flowing east at a constant speed of 1.40 km/h, and you paddle your kayak at a constant velocity, relative to the water, of 5.20 km/h at an angle of 15.0° east of north.

    (a) What is your velocity relative to a fixed point on the shore? Your velocity relative to the shore is: _________ km/h, at an angle of __________° east of north.

    (b) How much time, in seconds, does it take you to paddle from one bank of the river to the other, under the conditions described above? Note that you will not land on the far side at the point directly across the river from where you started.

    (c) While you are crossing the river, how far downriver do you travel (in meters)?

    IMG_3089.JPG IMG_3090.JPG
     
    Last edited: Feb 13, 2017
  2. jcsd
  3. Feb 13, 2017 #2

    jedishrfu

    Staff: Mentor

    So the river flows eastward and you must travel northward but not directly north you must travel 15 degrees east of north.

    So vacross = vkayak*cos(15) right?

    And vdown = vkayak*sin(15) + vriver

    Don't forget the degree / radian units setting must be degrees not radians if you're using a calculator.
     
  4. Feb 13, 2017 #3
    Yes, I used degrees. And I believe I used the components you listed too. Did I add them correctly in part A?
     
  5. Feb 13, 2017 #4
    To find the time to travel between both banks you could just find the y-component of the velocity and divide the distance between both banks by the result. Then using that time and the sum of both x-components, find the distance traveled downstream.
     
  6. Feb 13, 2017 #5
    But, dosen't the problem say that the kayak won't land directly across from the point you started from. Why do you suggest dividing the y-component by the distance between two banks in order to find the time?
     
  7. Feb 13, 2017 #6

    jedishrfu

    Staff: Mentor

    Vacross and vdown describe your kayaks actual motion as it tries to cross and is pushed down the river so vacross is the speed you're crossing the river

    480 = vacross x timetocross. Solve for timetocross

    Once you have time then you can determine how far down your kayak has drifted using vdown
     
  8. Feb 13, 2017 #7

    jedishrfu

    Staff: Mentor

    The angle isn't right it's 15 degrees east of north i.e. 15 degrees east of a perpendicular line across the river.

    I think you need a drawing of the river flowing east and the kayak launching 15 degrees east of north instead of this x,y stuff you have unless you state y means north and x means east.
     
  9. Feb 14, 2017 #8
    If I understand correctly, you're saying that you will drift as you move toward the other bank.

    If you find the y-component of your velocity then you're finding your northward velocity (irrelevant to the downstream (eastward) movement in this question).
     
  10. Feb 14, 2017 #9
    sorry, I'm a bit confused. how would I set up my vector components? The angle would not be 15 degrees?
     
  11. Feb 14, 2017 #10
    I'm
    The vector should point 15 degrees east of north, not north of east.
     
  12. Feb 14, 2017 #11
    Ah, ok, thank you. I think I understand (hopefully).

    are my answers correct now?

    a.) 5.02 km/h at 15 degrees E of N
    b.) 344 seconds
    c.) 134 meters
     
  13. Feb 14, 2017 #12
    I found the y-component like you suggested
     
  14. Feb 14, 2017 #13
    sorry, the angle should probably be different. 90 degrees since its the y-component going straight across (north)
     
  15. Feb 14, 2017 #14
    a) is 90 degrees east of north and the sum of both x-components.

    c) is 262 meters (I think you may have used wrong units here)
     
    Last edited: Feb 14, 2017
  16. Feb 14, 2017 #15

    jedishrfu

    Staff: Mentor

    I agree with a) and b) but c) doesn't look right

    c) = vkayak*sin(15) + vriver = 1.34 + 1.40 = 2.74 km/hr

    and your drift downstream = 2.74 * timetocrossinhours = X km
     
  17. Feb 14, 2017 #16
    Ok, gotcha. Thanks. Is the answer approximately 263 meters?
     
  18. Feb 14, 2017 #17
    Ok. Thanks for the help :) I redid the problem with your suggestion in mind and I got the right answer.
     
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