1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Physics Particle Movement Calculations

  1. Jan 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle moves in such a way that its position z, in meters, is given as a function of time t, by the equation z = 2t^2 − 3t^3. At what times is the particle at position z = 0? (two answers; one is 6E-1)


    2. Relevant equations
    z = 2t^2 − 3t^3



    3. The attempt at a solution
    I've tried plugging in certain numbers and i'm trying to figure out what my teacher means by 6E-1. Refreshing my memory of calculus etc..
     
  2. jcsd
  3. Jan 12, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    6E-1 means 6 x 10 to the power -1, which is 0.6.
    You can put 6E-1 into your calculator, press ENTER or = and see that.
    I don't think 6E-1 is a correct solution of that equation. Better check the question again.

    You have a cubic equation to solve, luckily an easy one.
    The usual approach is to factor it. Look for a common factor. For example, if you had this equation: 0 = 3x - 9, you would think as follows:
    The factors of 3x are 3 and x.
    The factors of the second term, 9, are 3 and 3.
    The factor in both is 3. So you would write down the 3, put brackets and ask yourself what 3 must be multiplied by to get the original 3x - 9:

    0 = 3x - 9 = 3(x - 3)
    To solve it, you ask if each factor could be zero. 3 can't be zero, but x-3 can.
    x - 3 = 0
    x = 3 (from adding 3 to both sides)

    example 2: 0 = 2t^3 + t^2 - t
    t is a common factor of all 3 terms. So
    0 = t(2t^2 + t - 1)
    A method called "trinomial factoring" can be used on the factor in the brackets.
    It is a bit too complicated to write easily in plain text, so I'll leave you to look it up.
    0 = t(t+1)(2t-1)
    Then t = 0 or t+1 = 0 or 2t-1 = 0
    and the 3 solutions are t = 0, t = -1 and t = 1/2
     
  4. Jan 13, 2009 #3
    Thank you very much for your help!

    I came out with the following conclusion:

    factored z=2t^2-3t^3 into z=t^2(2-3t)

    so I concluded that one answer is t=0

    because 0^2(anything) would = 0

    I also thought 2-3(.666666forever don't know how to write that on comp) would also equal 0 because 3*0.6forever would equal 2 so 0.6 was close.

    Is my thinking right?

    Thanks again for your help very much appreciated.
     
  5. Jan 13, 2009 #4
    Ok, new question related to this one:

    Use Differentiation to find an expression for the velocity of the particle as a function of time.

    I can't figure out what Differentiation is any ideas

    Thanks
     
  6. Jan 13, 2009 #5

    If your function was 2t^2-3t^3 you can get t^2(2-3t)=0, so t=o and then:

    2-3t=0 ---> -3t=-2 ---> t=(2/3) = .6666 (your teachers answer of 6E-1 i guess).

    As far as differentiation, when you differentiate a position function, you get a velocity function. When you differentiate a velocity function

    Im assuming you didnt learn differentiation yet? All you have to know for this question is the power rule:

    say you have ax^n, to differentiate put it in the form of nax^n-1...

    So heres an example:

    7x^5 ---->after differentiation you get 35x^4
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Physics Particle Movement Calculations
  1. Particle Movement (Replies: 10)

  2. Particle Movement (Replies: 4)

Loading...