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What are the components of the particle's velocity

  1. Sep 26, 2007 #1
    Two forces, F1 = (-6i - 4j)N and F2 = (-3i, 7j)N, act on a particle of mass 2 kg that is initially at rest at coordinates (-2.00m, +4.00m).

    (a) what are the components of the particle's velocity at t = 10s

    F1 = (-6i-4j)N
    F2 = (-3i + 7j)N

    Ff = F1 + F2

    Ff = (-9i 3j)

    F = ma; -9 = 2a = ai = -4.5m/s^2
    F = ma; 3 = 2a = aj = 1.5m/s^2

    V = -4.5(10)i + 1.5(10)j

    V = 45i + 15j

    (b) In what direction is the particle moving at t = 10s

    Vector F = sqrt ( (-9)^2 + 3^2) = root(90)

    tetha = arctan (3/-9)

    (c) what displacement does the particle undergo during the first 10s

    Xf = Xi + VixT + 1/2AxT^2

    deltaX = 0+ 1/2(9/2)(10)^2

    deltaX = 225m

    Yf = Yi + ViyT + 1/2AyT^2

    delta = 0 + 1/2(3/2)(10)^2 = 75m

    Are my answers right?
     
  2. jcsd
  3. Sep 26, 2007 #2

    Dick

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    Ok for a), except shouldn't it be -45i+15j. Ok for b). In c) check on signs for your displacement vector. It's moving down in x and up in y.
     
  4. Sep 26, 2007 #3
    oh, ok. So delta Y is -75m?
     
  5. Sep 26, 2007 #4

    Dick

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    Uh, "down in x, up in y"? Shouldn't delta X be negative? Feel free to disagree.
     
  6. Sep 26, 2007 #5
    oh, ya. my bad. You are right, dick.
    V = -45i + 15j

    and ya, therefore x is negative.
     
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