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Motion and vectors

  1. Feb 19, 2004 #1
    A particle moves is the xy plane with constant acceleration. At t = 0 the particle is at
    r1= (4m)i + (3m)j, with velocity v1. At t = 2s the particle has moved to r2= (10m)i- (2m)j and its velocity has changed to v2 = (5m/s)i – (6m/s)j. a) Find v1. 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?

    Since the acceleration is constanst can I use Vav =1/2(v0+v1)?
    Delta ri = ri2 - ri1
    = ((10m)i -(4m)i) = (6m)i
    Delat rj = rj2 - rj1 = (-2m)j - (3m)j = (-5m)j

    Thus Delta r = (6m)i - (5m)j

    and V(av)i = (6m)i/2s = (3m/s)i
    V(av)j = (-5m)j/2s = (-2.5m/s)j
    Thus V(av) = (3m/s)i - (2.5m/s)j

    and since Vav =1/2(v0+v1)
    we have (3m/s)i - (2.5m/s)j = 1/2(v0 + (5m/s)i – (6m/s)j)
    (3m/s)i - (2.5m/s)j = ((v0)/2) + (2.5m/s)i - (3m/s)j
    so v0 = 2(0.5m/s)i + (0.5m/s)j = (1m/s)i + (1m/s)j
    is that a logical reasoning for geting v?
    And by the way how can get the velocity of the particle as a function of time, and the position vector of the particle as a function of time?

    I really appreciate your help in advance.
    Thank you
     
  2. jcsd
  3. Feb 19, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Yes.
    Find the components of the acceleration, then use the equations for constant acceleration:
    Vx = Vx0 + axt, etc..
     
  4. Feb 21, 2004 #3

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    Sometimes applying Calculus is useful.
     
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