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Particle Movement

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle moves so that its position (in meters) as a function of time (in seconds) is R= ln(1+t)i+ (et-1)j + tk . Write expressions for (a) x-, (b) y- and (c) z-components of its velocity as well as the (d) x-, (e) y- and (f) z-components of its acceleration as functions of time.


    2. Relevant equations



    3. The attempt at a solution

    x-component= 1/1+t
    y-component= et
    z-component= 1

    How do I find their components of acceleration as a function of time??

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 15, 2009 #2

    kuruman

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    Take the derivative of a particular position component with respect to time twice. For example

    [tex]v_{x}=\frac{dx}{dt}[/tex]

    [tex]a_{x}=\frac{dv_{x}}{dt}[/tex]
     
  4. Sep 15, 2009 #3
    Ok so for the x-component 1/1+t...the derivative of that would be ....?? I am not good with fractions..Once I have that, I put it over 2t???
     
  5. Sep 15, 2009 #4

    kuruman

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    Write the fraction as (1 + t)-1, then take its derivative the usual way.
     
  6. Sep 15, 2009 #5
    so....-(1+t)^-2???

    then the next one will just say e^t?
    And the final one is 0??

    I found that derivative, but what do you mean with the twice time? Or is my work done?
     
  7. Sep 15, 2009 #6

    kuruman

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    You have found that vx=-(1+t)-2. To find the acceleration, you need to take the derivative one more time. So, to get the acceleration, you need to take the derivative two times, one to get the velocity and one more to get the acceleration. Once you get that, your work is not done yet. You need to repeat for the y and z components of the position.
     
  8. Sep 16, 2009 #7
    Ok I put it for the x ..2(1+t)^-3....and it was wrong..That is the second derivative though?
     
  9. Sep 16, 2009 #8

    kuruman

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    That is the second derivative of x, i.e. the x-component of the acceleration. What exactly do you mean "put it for the x"? What was the question that you used it as an answer to?
     
  10. Sep 16, 2009 #9
    I used that answer for part D, the x-component of acceleration.
     
  11. Sep 16, 2009 #10

    kuruman

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    I am sorry I misled you. :redface: I looked at what you had written down

    x-component = 1/(1+t)

    and I assumed that this was the x-component of the position. It is actually the velocity component. So you need to take the derivative of that to get the acceleration, which in this case gives you ax=-(1+t)-2. I apologize for the confusion it might have caused you. Your work was actually done when you asked if it was.
     
  12. Sep 16, 2009 #11
    No problem! I put the first derivatives in as the answers, and its right! Thanks for the help, and don't worry about the confusion, its ok!
     
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