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Position and Velocity Vectors

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle moves in the xy plane with constant acceleration. At t = 0 the particle is at rvec1 = (3.7 m)i + (3.4 m)j, with velocity vvec1. At t = 3 s, the particle has moved to rvec2 = (9 m)i − (1.9 m)j and its velocity has changed to vvec2 = (5.4 m/s)i − (6.5 m/s)j.


    (a) Find vvec1.


    (b) What is the acceleration of the particle?


    (c) What is the velocity of the particle as a function of time?


    (d) What is the position vector of the particle as a function of time?

    I need some guidance. I tried using the average velocity formula for (a), but it doesn't seem to be working for me. I tried doing 9-3.7 / 3. for the i vector. And likewise, for the j vector (-1.9-3.4)/3. I get 1.76i-1.76j, and it isn't correct.

    I need some guidance on the others too. Thanks guys.
     
  2. jcsd
  3. Oct 13, 2009 #2
    Jeez, I cannot seem to get this! I have been trying to get it all night.
     
  4. Oct 13, 2009 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Loppyfoot! :smile:

    (use bold for vectors :wink:)
    Distance velocity and acceleration are all vectors, and so they add like vectors.

    Try s = vt + (1/2)at2 :wink:
     
  5. Oct 14, 2009 #4
    How should I implement that equation into part a), if I do not know the acceleration vectors yet?
     
  6. Oct 14, 2009 #5

    Mark44

    Staff: Mentor

    Part of your problem description doesn't make sense to me.
    Edit: Now it does make sense.
    Edit: Ignore the following.
    This description says that the particle moves with constant acceleration. Later it says that the velocity has changed to ... If the acceleration is constant, the velocity can't change, since acceleration is the instantaneous rate of change of velocity with respect to time.
     
    Last edited: Oct 14, 2009
  7. Oct 14, 2009 #6
    If the acceleration is constant, doesn't that mean that the velocity is changing at a constant rate?
     
  8. Oct 14, 2009 #7

    Mark44

    Staff: Mentor

    Never mind. I take back what I said. I was thinking zero acceleration, not constant acceleration.
     
  9. Oct 14, 2009 #8
    I think I figured it out. I applied one of the 4 kinematic equations, and then I should be able to get the rest from there.

    EDIT: I got a and b, but how would I go about getting c and d?
     
  10. Oct 14, 2009 #9

    Mark44

    Staff: Mentor

    For c, if you got b, use it to get the velocity. a = dv/dt, so you can integrate what you have for a to get v as a function of t. You'll get a constant (vector) of integration, but you know v(3), so should be able to figure out the constant.

    For d, do essentially the same thing: v = ds/dt. Integrate that to get s and use the given information about s(0) to figure out this constant (vector) of integration. Does that make sense?
     
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