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Homework Help: Vectors in 2D

  1. Sep 15, 2014 #1
    1. The problem statement, all variables and given/known data
    A canoe has a velocity of 0.360m/s southeast (at 315°) relative to the earth. The canoe is on a river that is flowing at 0.580m/s east relative to the earth.

    Find the magnitude of the velocity v⃗ c/r of the canoe relative to the river.

    2. Relevant equations


    3. The attempt at a solution

    The answer is Vc/r=0.413 m/s and the (Vc/r)x , (Vc/r)y =
    -0.325,-0.255 ; respectively.

    There are a few things I am confused about with the equations on the problem in general. Firstly, why do these equations work : Vx=V0cos(θ) and Vy=V0sin(θ).

    I don't understand why I have to use these equations in the problem to obtain the Vx when the problem (to my understanding) gives me Vx and the Vy, which should be the value of Vc/r since this should be just simple vector addition. This is what I am thinking (see attachments). The magnitude of the velocity of Vc/r = Vc/e - Vr/e. So, Vc/e is given at a 315° angle at a velocity of .350m/s. Then using the head to tail method, -Vr/e is given at 180° at a velocity of .580m/s. Using the cos(315°)*.360, I should get -.255 for the magnitude of the velocity of Vc/r instead of .413. I know this is obviously wrong since you can't have a negative velocity in this problem, the way they have the positive and negative planes set up, but I just want to know why the other way works and why it isn't simple vector addition/subtraction. Thanks in advance!

    Attached Files:

  2. jcsd
  3. Sep 15, 2014 #2


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    The velocity of the canoe relative to the earth is the resultant vector of the canoe's velocity relative to the river and of the velocity of the river relative to the earth.
  4. Sep 15, 2014 #3
    That doesn't really help me, could you explain more in depth please?
  5. Sep 15, 2014 #4

    rude man

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    Gold Member

    vc/e = vr/e + vc/r

    Solve the equation for vc/r by breaking the other two (given) velocities into their x (east) and y (north) components, then take the magnitude of the result. The given answer checks.

    NOTE: the given velocity of the canoe w/r/t earth is not the velocity the canoe would have w/r/t earth in the absence of river flow. It's the actual canoe-to-earth velocity including the effect of the river flow.
  6. Sep 15, 2014 #5


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    The canoe's velocity consists of two components. Its velocity relative to the river, that is the person rowing it sees that he is rowing the canoe at a certain speed in a certain direction with respect to the water, but it is also carried by the river at a certain speed in a certain direction. So the canoe's velocity is the resultant of these two velocities.
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