1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Vector problem ?

  1. Sep 26, 2006 #1
    1. The points S and T are midpoints of the sides AB and AD respectively of a parallelogram ABCD .The line CS and CT cut the diagonal BD at the points U and V respectively .
    Show that BU = x BC + x CD and also BU = (1-y)BC +1/2 y CD , where x and y are constant.Hence show that BU=1/3 BD
    Deduce that the lines CS and CT trisect the diagonal BD .

    I can do the first two part but I don't know how to deduce that the lines CS and CT trisect the diagonal BD .Please help me .

    2. A canal of width 2a has parallel straight banks and the water flows due north. The points A and B are on opposite banks and B is due east of A , with point O as the midpoint of AB . The x-axis and y-axis are taken in the east and north directions respectively with O as the origin. The speed of the current in the canal ,Vc is given by
    Vc=Vo(1-x^2/a^2) where Vo is the speed of the current in the middle of the canal and x is the distance eastwards from the middle of the canal.A swimmer swims from A towards east at speed Vr relative to the current in the canal . Taking y to denote the distance northwards travelled by the sewimmer ,show that
    dy/dx = Vo/Vr (1-x^2/a^2) .
    I have no idea with this question . Is it want to find the velocity of the swimmer relative to ground ? But I cannot find the answer also if I find the velocity of the swimmer relative to ground . How to do this ?

    Please help me . Thanks .:!!)
  2. jcsd
  3. Sep 26, 2006 #2


    User Avatar
    Science Advisor

    1. You are able to show that BU= 1/3 BD? That says immediately that U is 1/3 the distance from B to D. You should by exactly the same argument, using D instead of B, be able to show that DV= 1/3 BD. That is, U and V cut BD into thirds- they trisect BD.

    2. The speed of the swimmer, relative to the water, is Vr i and the speed of the current is V0(1-x^2/a^2)j so the speed of the swimmer relative to the ground is the sum of those vectors, Vr i + V0(1-x^2/a^2)j.
    That is [itex]\frac{dx}{dt}= V_r[/itex] also
    [tex]\frac{dy}{dt}= V_0(1- \frac{x^2}{a^2})[/tex]
    Now, just use the chain rule:
    [tex]\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Vector problem ?
  1. Vector Problem (Replies: 3)

  2. Vector problems (Replies: 8)

  3. Vector Problem (Replies: 13)

  4. A vectors problem (Replies: 14)

  5. Vector Problem (Replies: 15)