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Position vector of a moving particle (calc based intro)

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Reviewing for a midterm. I have this problem I can't seem to nail. Want to try it?

    "The acceleration of a particle moving only in a horizontal xy-plane is given by a=<(3t)i,(4t)j>, where 'a' is in m/s^2 and 't' is in seconds. At t=0, the position vector r=<(20)i,(40)j> locates the particle which then has the velocity vector v=<(5)i,(2)j>. At t=4s, what is the particle's position vector in unit-vector notation?"

    to clarify units: v is in m/s, a is in m/s^2, position is in meters.


    Answer: r=<72i,90.2j>


    2. Relevant equations

    The standard kinematics equations, I'm assuming. If calc is involved, I wouldn't know where to implement it.

    3. The attempt at a solution

    I think it's the acceleration at t=4 seconds that's tripping me up. When I plug in for position in the 'x' or 'y' direction I am using t=4 to multiply with the original acceleration components. Am I not finding acceleration correctly? And wouldn't it stand that if I multiplied an acceleration component by a time I would get a velocity?
     
  2. jcsd
  3. Mar 7, 2010 #2

    lightgrav

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    if you were given x(t) , how would you find a(t) ?
    yes, using calculus.
    this is the "inverse" question ; they give a(t) ... not constant ... and want x(t)
    ... so use the "inverse" operation.
     
  4. Mar 7, 2010 #3
    Yes, if you want to find acceleration from position you would take two derivatives right?

    So if you want to find position from acceleration you would try to "undo" derivatives (don't forget the constants).
     
  5. Mar 7, 2010 #4
    This is very helpful, thank you. Now, in general, would I consider integrating and/or differentiating in a situation where my acceleration isn't constant?

    I ask because I want to know if there is a way to do this problem with your 'plain-vanilla' kinematics equations.

    Actually, let me restate. When should I consider integration and or differentiation in a physics problem?
     
  6. Mar 7, 2010 #5

    lightgrav

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    Here, the acceleration increases at a constant rate (da/dt = jerk = const),
    so you _can_ find the average acceleration in a simple route , here .
    ... if you want to, you can sprinkle a " + 1/6 j t^3" term onto the vanilla
    ... and dip it in fudge by adding " + 1/24 d t^4 " , etc ... where to stop ?!?

    well, you got those "plain-vanilla" kinematic equations via time-derivatives of location .
    The better you understand things, the less you need to memorize;
    but some things are SO often encountered (like constant accel)
    and some derivations are TOO nasty of algebra to do on an Exam.
    . . . so, you should memorize a few results (2 per chapter?)
     
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