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Havin' problem in vector direction

  1. Oct 28, 2005 #1
    What i have now is the initial relative position of A from B, which is 5j
    initial velocity of A is 40i + 30j, the direction is changed so that it can overtake B.
    Velocity of B is 12i + 16j.
    I can find the speed of A, which is 50.
    I tried to let the new velocity of A be 50(ai + bj), then here's the problem:
    I tried to use
    [tex] \\vec{r}_A = \\vec{r}_B [/tex]
    r_A=r_B for which r represents the displacement.
    but there are 3 unknowns and i'm stuck there...
    *I'm having problem with the latex code... hope u can understand
    Last edited: Oct 28, 2005
  2. jcsd
  3. Oct 28, 2005 #2
    so what are you trying tofind
  4. Oct 28, 2005 #3
    I'm trying to find the new direction of velocity of A
  5. Oct 28, 2005 #4
    so do this

    a is at the orgin

    then b is at 0,5

    if the vel of b is 12i + 16j

    where would that put it in the next unit of time
  6. Oct 28, 2005 #5
    I'm sorry, the s should be located north of b,
    so b is the origin, and a at (0,5).
    If vel of b is 12i + 16j, then in next unit of time,
    it'll reach the point (12,16).
    What can it help me?
    I still dun understand.
  7. Oct 29, 2005 #6
    ok now what would be the vector be to get a to that point?
  8. Oct 29, 2005 #7
    12i + 11j?
    u mean for A?
    still dun understand.
  9. Oct 29, 2005 #8
    if you are going from 5j to 12i + 16j then the vector be 12i + 16j
  10. Oct 29, 2005 #9
    What can i get from vector 12i + 11j, what to do next?
    Sorry i'm too dumb... i need more explaination if possible...
  11. Oct 31, 2005 #10
    no one can help me?
  12. Oct 31, 2005 #11


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    Homework Helper

    What do you want to know? I don't understand this notion of 'to overtake B'. Do they mean to meet B (as in, calculating the angle to fire a bullet to hit a moving target)?
  13. Oct 31, 2005 #12


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    Homework Helper

    Well I'm guessing you want to know the new direction of A, such that A will meet B, like a bullet hitting a moving target.

    I drew an obtuse triangle with sides {5,2x,5x}, then used the cos-rule (one angle is known) to get an equation in x, which I solved. Knowing x, one knows the lengths and finding the angle is just a small step away.
    Last edited: Oct 31, 2005
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