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Objects moving in a plane.

  1. Sep 26, 2007 #1
    I'm having terrible difficulty starting this problem, it's one of the chapter excercises in the book and it's revisited in the homework later on. I'm going to give different data, as I would like to actually solve this one myself, I just need a kick start...

    1. The problem statement, all variables and given/known data
    We've got an object moving in a plane, no velocity is stated at all, just that it's moving. It's mass is 6.00 kg and it's coordinates are given by 2 equations, [tex]x = 4t^2 - 1[/tex] and [tex]y = 2t^3 + 6[/tex]. They are asking what the net force acting on this object is at time t = 5.00s.

    2. Relevant equations
    I know somewhere in there I'm going to use kinematic equations. I started by trying to find [tex]\Delta X[/tex] and [tex]\Delta Y[/tex]...

    Thanks to anyone who can point me in the right direction!
    Last edited: Sep 26, 2007
  2. jcsd
  3. Sep 26, 2007 #2


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    It's in a plane, so position, velocity, acceleration, and force are all vectors with x and y components. The position vector is (x,y)=(4t2-1,2t3+6). Can you find the velocity vector? (How is velocity related to position?) Then, can you find the acceleration vector? Then, can you find the force vector?
  4. Sep 26, 2007 #3
    Ah, that makes a lot of sense! Thank you so much! I was completely overlooking that.
  5. Sep 26, 2007 #4


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    Glad to help.
  6. Sep 26, 2007 #5
    If I'm using position to get a velocity vector with the formula [tex]V_x_{avg} = \frac{\Delta x}{\Delta t}[/tex], can I use t = 0 for my [tex]t_i[/tex]?
  7. Sep 26, 2007 #6


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    You should be computing instantaneous velocity, not average velocity.

    I assume this is a calculus-based course?
  8. Sep 26, 2007 #7
    For the formulas I have you still need [tex]\Delta x[/tex] and [tex]\Delta t[/tex]

    [tex]v_x = lim_{\Delta t \rightarrow 0}\frac{\Delta x}{\Delta t}[/tex]

    ...and yes, this is calculus based.

    I feel like I've missed a lesson or missed something in class.
  9. Sep 26, 2007 #8


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    That limit defines the derivative of x with respect to t. Given x as a simple function of t, say, x=t2, can you compute the derivative dx/dt ?
  10. Sep 26, 2007 #9
    OH! So it would be 2t then... If the function was in fact [tex]t^2[/tex].
  11. Sep 26, 2007 #10
    You would need two formulas:

    1. Newton's second law [itex] \mathbf{F} = m \mathbf{a} [/itex] and
    2. definition of components of the acceleration vector

    [tex] a_x = d^2x(t)/dt^2 [/tex]
    [tex] a_y = d^2y(t)/dt^2 [/tex]

  12. Sep 26, 2007 #11
    So, if my position in the x direction is a function of time, like [tex]x=2t^2[/tex] the derivative of that is [tex]4t[/tex] which should be my velocity in the x direction. Then a second derivative should give me 4 and that should be my acceleration in the x direction. Am I on the right track?
  13. Sep 26, 2007 #12
    Yes, you got it.

  14. Sep 26, 2007 #13
    Thanks! It's much appreciated. Good thing I have a whole week to finish studying for my test!
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