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1D motion

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data

    The position s (in metres) of a particle, as a function of time t (in seconds), is described by the equation s=5i+4r^2j. i and j are perpendicular unit vectors.
    a) Calculate the position of the particle at t=5.
    b) Determine the particle's average velocity between t=0 and t=2.
    c) Determine the particle's instantaneous velocity at t=4.


    3. The attempt at a solution

    a) When t=5,
    s=5i+4(25)j=5i+100j
    s=√10025=100m

    b) When t=0,
    s=5i=5m
    When t=2,
    s=5i+4(4)j=5i+16j=√281=16.76m

    average velocity=displacement/time=(16.76-5)/(2-0)=5.9m/s

    c) By differentiation,
    v=4(2)tj=8tj
    When t=4,
    v=8(4)tj=32j=32m/s
     
  2. jcsd
  3. May 7, 2014 #2
    Hi,

    The position of a particle is a vector right? √(52+1002) is the distance.

    Again, average velocity is a vector. Displacement is the vector difference of the final and initial positions.
     
  4. May 7, 2014 #3
    Thanks for your reply.

    Do you mean I have to give the direction as well? The angles?
    For a),
    tanθ=100/5, θ=1.5°
    Therefore, the position is 100m 1.5°

    And for b), tanβ=16/5, β=1.3o
    Therefore, the average velocity is 5.9m/s 1.3°

    Like these?
     
  5. May 7, 2014 #4
    Your answer is right for part (A) but not for part (b).

    In part (b), you are supposed to find [itex]\frac{\vec{s_2}-\vec{s_1}}{Δt}[/itex]. You just have to subtract each component (i.e. like final x-initial x and final y- initial y) separately and then divide the resultant vector by t. Your answer should look like [itex] \frac{(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}}{Δt}[/itex].
     
  6. May 7, 2014 #5
    You made a mistake on part b because yo wrote
    s=5i=5m
    Never ever do that.
    either something is a vector (5i is a vector), or something is a scalar (5m is a scalar). But something cannot be both a scalar and a vector at the same time, so the equation 5i=5m makes no sense. if s is a vector than you should denote that somehow. One common convention is to use boldface so your equation should look like this
    s=5i
    |s|=5m.

    Always make sure to do that (even on your scraps that you're going to throw away anyways).
     
  7. May 7, 2014 #6
    One more thing. This is a 2-D motion but the title of the thread says 1-D motion.
     
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