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Homework Help: 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,

    b) When t=0,
    When t=2,

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

    c) By differentiation,
    When t=4,
  2. jcsd
  3. May 7, 2014 #2

    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
    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

    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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