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Homework Help: Expression of Acceleration and velocity

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    The position of an object of mass m is given by:
    http://imageshack.us/photo/my-images/62/20110920161443.jpg/
    Find expression for the velocity and the acceleration and deduce the force components acting on the particle.
    2. Relevant equations
    W = vr, v = u+at , f = ma
    v = d/t

    3. The attempt at a solution
    I was thinking of finding out the magnitude of the position (by squaring them followed by square root),so then I could divide position by time to get velocity then use SUVAT equation but so far I have achieved nothing sensible.. I would like to know am I on the right track ?
     
    Last edited: Sep 20, 2011
  2. jcsd
  3. Sep 20, 2011 #2

    Mark44

    Staff: Mentor

    I don't think so.

    You have r as a (vector) function of t. v is defined as dr/dt, and a is defined as dv/dt.
     
    Last edited: Sep 20, 2011
  4. Sep 20, 2011 #3

    lanedance

    User Avatar
    Homework Helper

    you can just differentiate the components to find the velocity (once) and acceleration (twice) components

    I see no need for taking the magnitude, unless it asks for one

    deduce the force using f=ma in component form
     
  5. Sep 21, 2011 #4
    Thanks for your replies everyone!
    Hm.. if I have understood the above posts correct then is this working of mine , right ?
    r = (at+bt^2) i +4j + (Ce^-4t sin wt) k

    Right, so by differentiating each components separately once i.e the first derivative should give me vector function of velocity.

    component z becomes (at+bt^2) becomes: (t +2t) , 4J becomes J , and for the last component I use chain rule (un sure , I don't think product rule is appropriate)
    (ce^-4t sinwt) k = (-4te^-4t cos w ) k


    Am I right? I am slightly confused about the 'constants' .....
    hmm
     
  6. Sep 21, 2011 #5

    Mark44

    Staff: Mentor

    Most of what you wrote is incorrect.
    For the x (not z) component, d/dt(at + bt2) [itex]\neq[/itex] t + 2t.
    For the y component, d/dt(4) [itex]\neq[/itex] 1.
    For the z component, you need to find d/dt(Ce-4tsin(wt)). To do that, you will need to use the product rule, and then the chain rule (twice).
     
    Last edited: Sep 21, 2011
  7. Sep 21, 2011 #6
    Thanks for the reply, could you tell me what exactly happens to the constants ?
     
    Last edited by a moderator: Sep 21, 2011
  8. Sep 21, 2011 #7

    Mark44

    Staff: Mentor

    d/dt(K) = 0 and d/dt(a*f(t)) = a * f'(t)
     
  9. Sep 21, 2011 #8
    Sorry but I am further lost with the above expression.
    Oh as for the derivative of 4j goes.. wouldn't that be 4 and a+2bt for i , so do the constants stay as they are ?
    I think that makes sense since a 'constant' is any integer... for example if a was replaced by 3t then the derivative of that in respect to time would have been 3.
     
    Last edited: Sep 21, 2011
  10. Sep 21, 2011 #9

    Mark44

    Staff: Mentor

    The first one says that the derivative of a constant is zero.

    The second one says that the derivative of a constant times a function is the constant times the derivative of the function.

    If you are finding derivatives of vector-valued functions, and you don't know these differentiation rules, you are in big trouble!

    Look in your book at some examples of differentiating vector-valued functions. You don't actually differentiate the unit vectors i, j, and k. d/dt(4) [itex]\neq[/itex] 4. For the other one, yes, d/dt(at + bt2) = a + 2bt
     
  11. Sep 21, 2011 #10

    Mark44

    Staff: Mentor

    No, a constant is a fixed number. It could be an integer, but it doesn't have to be. For example, in the expression [itex]\pi x[/itex], [itex]\pi[/itex] is a constant, but it isn't an integer.
    But a is a constant, so it doesn't depend on any variable. That's why it's constant.
     
  12. Sep 21, 2011 #11
    Thanks that makes much sense now.. sorry , I thought f and a above were force and acceleration.
    Oh I see.. yea how stupid can I be... ijk denote Axis... That was very sloppy of me...
     
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