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Solving position vectors involving velocities and time

  1. Sep 2, 2010 #1
    A particle P has velocity (-3i + j)m/s at time t=0. The particle moves with constant acceleration a = (2i+3j)m/(ss). Find the speed of the particle after 3 seconds.

    I've got no idea how to solve this. Please help.
     
  2. jcsd
  3. Sep 2, 2010 #2

    Mark44

    Staff: Mentor

    Integrate a with respect to t to get v. For the speed when t = 3, find the magnitude v(3).
     
  4. Sep 2, 2010 #3

    berkeman

    User Avatar

    Staff: Mentor

    Acceleration is defined as the change in velocity over some time period, or

    [tex]a = \frac{\Delta v}{\Delta t} [/tex]

    You have an initial velocity vector. The acceleration for that amount of time will change the velocity vector like this:

    [tex]v(t) = v_0 + \Delta v = v_0 + a \Delta t[/tex]

    Does that help?


    EDIT -- Mark44 beat me to it (again)!
     
  5. Sep 2, 2010 #4
    Wow. How do you integrate a vector?
     
  6. Sep 2, 2010 #5

    Mark44

    Staff: Mentor

    Integrate component-wise.

    Even though you posted in the Precalc section, this seems to be a calculus-based problem to me, since you're being asked to find the velocity vector given the position vector.
     
  7. Sep 2, 2010 #6
    hmm.
    I've got the steps from the book here, but I dont understand what theyre doing. Did they integrate here too?
     

    Attached Files:

  8. Sep 2, 2010 #7

    Mark44

    Staff: Mentor

    No, they're not integrating. In the problem you attached, velocity is constant, so they get a unit vector in the direction of the velocity, and then multiply by the velocity to get the position. For example, if you are driving at a constant velocity in a car at 60 mi/hr, then your position after 1 hour, relative to where you started, is 1 hr * 60 mi/hr = 60 mi. That's a little like what they did.

    For your problem, the acceleration is constant, so find a unit vector with the same direction as the acceleration, then multiply by 3 (sec). That will give you the speed at t = 3 sec.

    Follow the basic idea given in the thumbnail you attached, only where they talk about r and v, you'll be working with v and a. Does that make sense?
     
  9. Sep 3, 2010 #8
    Oh wow. I'm stupid. I posted asking help for the wrong question .
    The question in the first post was easy.
    It's the question in the thumbnail I don't understand, and I'm totally confused.
    I think the question I should be asking is, what is the magnitude of a vector, and how do you get a unit vector.
    I'm sorry for asking so much, but you see I've only just started vectors.

    EDIT: I did some wiki-ing and now understand a little more, but what I don't get is why multiplying the unit vector by the speed gives you the velocity.

    EDIT 2: Ok, so the magnitude of a vector is the length of the vector. The magnitude of velocity is the speed. So the length of the vector is the speed. The unit vector has a magnitude/length of one. So the vector divded by the length is the unit vector, meaning the velocity divded by the speed is the unit, giving the speed times the unit is the velocity. *phew*
    But then. If the speed is the magnitude of the vector, and the vector is 3i-4j, then the magnitude/speed of the vector would be 5 units. But if the speed is 5 units, it's not 15m/s ....... unless (I've just thought of this now as I type).. a unit is three meters... have I understood this correct?
     
    Last edited: Sep 3, 2010
  10. Sep 3, 2010 #9

    Mark44

    Staff: Mentor

    In the problem in the attachment, the position vector is r = 3i - 4j. The magnitude of this vector, |r|, is 5. This isn't the speed - it's just the magnitude of the position vector. They are finding the magnitude of the position vector to come up with a unit vector in the same direction as the position vector.

    This unit vector is (1/5)(3i - 4j) = (3/5)i - (4/5)j.

    To get the velocity vector, they multiply the unit vector above by 15, getting 15((3/5)i - (4/5)j) = 9i - 12j. The magnitude of the velocity vector (the speed) is sqrt(9^2 + 12^2) = sqrt(81 + 144) = sqrt(225) = 15 (m/sec), the same as was given.
     
  11. Sep 3, 2010 #10
    Ok I understand that, but what I don't see is why you multiply the speed with the unit vector to get the velocity.
     
  12. Sep 3, 2010 #11

    Mark44

    Staff: Mentor

    To get a vector you need its direction and magnitude. The unit vector gives you the direction of the velocity vector. The speed gives you the magnitude of the velocity vector.
     
  13. Sep 3, 2010 #12
    Aaah, I get it now. Thanks
     
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