Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Solving position vectors using given speed.

  1. Sep 3, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle P starts at the point with position vector 4i + j. P moves with constant velocity vm/s. After t seconds, P is at the point with position vector 12i - 11j. Find t if the speed of P is 4m/s.

    2. Relevant equations
    n/a


    3. The attempt at a solution
    6i - 3j = -2i + 3j + vt
    => 8i - 6j = vt
    => 8i - 6j = *bleh. and now I'm stuck.*
     
  2. jcsd
  3. Sep 3, 2010 #2

    Mark44

    Staff: Mentor

    Use the formula in the attachment in your other problem, r = r0 + vt.

    In your attempt, where did 6i - 3j and -2i + 3j come from? They don't have anything to do with this problem.

    At 0 seconds, the particle is at r0 = 4i + j. After t seconds, r = 12i - 11j.

    So 12i - 11j = 4i + j + v t.

    Start with this equation, simplify it a bit, and use the idea that if two vectors are equal, their magnitudes are also equal.
     
  4. Sep 3, 2010 #3
    HAHAHAHA. This is the second time I've done this. *facepalm* I copied the wrong vectors.
    r0 = -2i + 3j, and r = 6i - 3j.

    So, 8i - 6j = tv

    And I really have no idea what to do to proceed.

    EDIT: Ok. So,
    8i - 6j = vt
    Both sides are equivalent.
    Speed = |v|
    So,
    | 8i - 6j | = |vt|
    => 10 = 4t
    => t= 2.5
    So t=2.5 seconds?
    But can you just multiply like that?
    What I mean to say is that is |vt| = |v| * t ?
     
    Last edited: Sep 3, 2010
  5. Sep 3, 2010 #4

    Mark44

    Staff: Mentor

    Both sides are equal. There is a difference. Statements can be equivalent (same truth values); expressions can be equal (or less than, greater than, etc.).
    Yes, the magnitude of a scalar times a vector is the scalar times the magnitude of the vector.

    More precisely, |kv| = |k||v|. This takes into account the possibility that k is a negative number.
     
  6. Sep 3, 2010 #5
    Wait, don't you mean |kv| = k * |v|
    Or are you just implying that the magnitude of a scalar is the sacalr itself, i.e, |k| = k

    *sigh* I need some sleep.
     
  7. Sep 3, 2010 #6
    |k| means the absolute value of the scalar k, which is always positive, like the magnitude (length) of a vector. So if k = -5, then |k| = 5. If k = 5, |k| = 5. It just means, throw away the minus sign if there is one.
     
  8. Sep 3, 2010 #7

    Mark44

    Staff: Mentor

    No, I mean exactly what I wrote, namely that |kv| = |k||v|. The magnitude of a scalar is its distance from 0.

    For example, let v = 3i + 4j, so |v| = 5.
    |-2v| = |-2(3i + 4j)| = |-6i - 8j)| = [itex]\sqrt[/itex]((-6)2 + (-8)2) = [itex]\sqrt[/itex](36 + 64) = [itex]\sqrt[/itex](100) = 10 = 2|v|
     
  9. Sep 3, 2010 #8
    Oh.
    I can't wait till I'm as smart as you guys. *anticipation emoticon*
    EDIT: And thanks for all your help.
     
    Last edited: Sep 3, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook