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Homework Help: Particle at rest

  1. Jan 17, 2014 #1
    1. The problem statement, all variables and given/known data

    A particle's position is r =(ct^2−2dt^3) i hat+(2ct^2−dt^3)j hat. where c and d are positive constants.
    part a) Find expressions for times t > 0 when the particle is moving in the x-direction.
    part b) Find expressions for times t > 0 when the particle is moving in the y-direction.

    Is there any time,t >0 when the particle is c) at rest and d) accelerating in the x-direction?
    If either answer us "yes", find the time(s).

    3. The attempt at a solution


    r''(t)→ = d^2/dt^2 [ ct^2 -2dt^3] i
    a(t)→ = 2c - 12dt
    t = (2c - a)/12d


    r"(t)→d^2/dt^2[ 2ct^2 - dt^3] j
    a(t)→ = (4c -6dt)j
    t = (4c - a)/6d


    If particle is at rest, then velocity = 0

    v→ = r'(t)→ = (2ct - 6dt^2) i + (4ct - 3dt^2) j
    0 = (2ct - 6dt^2) i + (4ct - 3dt^2) j

    how do I proceed from here?
  2. jcsd
  3. Jan 17, 2014 #2
    For (a) and (b), what does "moving in x or y direction" mean?

    For (c), find t such that the equation is satisfied.
  4. Jan 17, 2014 #3
    I suppose it means (x,0) and (0,y). Word for word from the book.
    Part (a) and (b) is correct. How should I attempt part(c)?
    I set dr/dt to be zero but I can't reduce the expression.
  5. Jan 17, 2014 #4
    How do you know that (a) and (b) are correct? I have a very different interpretation of those parts.

    For (c), you have a vector equation. That means you have equations for each component. Find a common solution for the component equations, if it exists.
  6. Jan 17, 2014 #5

    It seems to make sense, although, I would rule out my answer being a probable falsity.
    Let's begin from part(a) then.
    Last edited: Jan 17, 2014
  7. Jan 17, 2014 #6
    When you hear "something is moving in some direction", is this a statement about the position of something, or its velocity, or its acceleration?
  8. Jan 17, 2014 #7
    Position. But this change in position could further imply the velocity is changing, and, therefore, a change in the acceleration too.
  9. Jan 17, 2014 #8
    So if you are at x = 5 m and y = 6 m, which direction are you moving?
  10. Jan 17, 2014 #9
    50° from the origin-origin being (0,0)

    Edit: It cannot be ascertain.
  11. Jan 17, 2014 #10
    Even if you are standing still at that position?
  12. Jan 17, 2014 #11
    I edited the post you responded to.
    It cannot be ascertain unless the change in position is made relative to another position, otherwise, the question becomes meaningless. Is that what you were trying to expound?
  13. Jan 17, 2014 #12
    Yes. And how do you characterize "change in position"?
  14. Jan 17, 2014 #13
    Δs = vit + 0.5at^2
  15. Jan 17, 2014 #14
    That is valid only for uniformly accelerated motion. Imagine ##\Delta t## is very very small. How can you approximate ##\Delta s##?
  16. Jan 17, 2014 #15
    change in position/ change in time = velocity.

    v(t) . Δt = Δs = sf - si

    or in other words; the derivative of position x change in time
  17. Jan 17, 2014 #16
    Here's another go:


    If we are concern only with the velocity of the particle along the x-axis, then, this implies that the velocity along the y-axis is hypothetically = 0ms^-1

    v = (vx=0,vy)
    vy = 0ms^-1
    vy = (dr/dt)y = 0ms^-1
    ry→ = (2ct^2 -dt^3)j
    dr/dt = 4ct - 3dt^2
    4ct - 3dt^2 = 0
    4c = 3dt
    t = 4c/3d


    v = (vx,vy=0)
    vx = 0ms^-1
    vx = (dr/dt)x = 0ms^-1
    rx→ = (ct^2 -2dt^3)i
    (dr/dt)x = (2ct -6dt^2)i
    2ct = 6dt^2
    2c = 6dt
    t = 2c/6d =c/3d


    if the particle is at rest, that implies that v = (vx = 0, vy = 0)

    at t = 4c/3d, vy = 0
    at t = c/3d, vx = 0

    there is no time at which tx = ty, so, the particle is never at rest.


    particle accelerating only in the x-axis implies it's acceleration along the y-axis = 0

    py'(t)→ = 4ct - 3dt^2
    py"(t)→ = a(t)→ = 4c - 6dt

    4c = 6dt
    t = 2c/3d
    Last edited: Jan 17, 2014
  18. Jan 18, 2014 #17
    Your method and results look good to me.
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