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Homework Help: Particle Position and Time

  1. Sep 7, 2008 #1
    1. The problem statement, all variables and given/known data
    he position of a particle moving along the x axis depends on the time according to the equation x = ct2 - bt3, where x is in meters and t in seconds.

    (a) What dimension and units must c have?
    m/s2
    s2/m
    m2/s
    s/m2


    What dimension and units must b have?
    s/m3
    m3/s
    s3/m
    m/s3


    For the following, let the numerical values of c and b be 3.2 and 1.0 respectively.

    (b) At what time does the particle reach its maximum positive x position?


    (c) What distance does the particle cover in the first 4.0 s?


    (d) What is its displacement from t = 0 to t = 4.0 s?


    (e) What is its velocity at t = 1.0?

    What is its velocity at t = 2.0?

    What is its velocity at t = 3.0?

    What is its velocity at t = 4.0 s?


    (f) What is its acceleration at at t = 1.0 s?

    What is its acceleration at at t = 2.0 s?

    What is its acceleration at at t = 3.0 s?

    What is its acceleration at at t = 4.0 s?



    2. Relevant equations

    I know that in the bold are correct. Unless of course I'm proven to be wrong on this of course.

    So, that seems to use the grand dx/dt for (b) unless I'm incorrect. Though, I'm not sure entirely.

    So, it would be as such unless I'm mistaken.

    v = dx / dt = [(3)*35 + (2) 1]=0

    Though, something tells me that I may be incorrect, if so can you please explain why in details. (Sorry to be demanding.) I feel it has to do with the coefficient.
     
  2. jcsd
  3. Sep 7, 2008 #2
    For a) your 1st answer is incorrect, but the second is correct, assuming the numbers 2 and 3 are exponents.

    For b), yes, set dx/dt = 0 and solve for t; I don't understand your answer.

    x = ct2 - bt3

    [tex]\frac{dx}{dt}[/tex] = 2ct - 3bt2 = 0,

    etc.
     
  4. Sep 7, 2008 #3
    I noticed that I made a mistake and its m/s^2. Thanks for that, though.

    Ahh, I noticed I have to move the coefficient down and lessen it by one for the reminder. I presume you would then just fill in with the two numbers given. Am I doing this wrong so far?

    Next you would get two derivatives, right?

    So it would be a quadratic function, though am I missing anything?
     
  5. Sep 7, 2008 #4
    I don't understand your question.
     
  6. Sep 7, 2008 #5
    Sorry.

    Once you get: dx/dt=2(3.2)t-3(1)t^2=0 you would then find the derivative, right or not? Which would be a quadratic function. If so you would then solve that.
     
  7. Sep 7, 2008 #6
    dx/dt=2(3.2)t-3(1)t^2 is the derivative. Set it = to 0 and solve.
     
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