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A moving particle question

  1. Mar 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle moves so that s=108t-t^4, commencing at time t=0 seconds.
    a) during what time interval is its velocity positive?
    b) during what time interval is its acceleration negative?

    2. Relevant equations
    t=0

    3. The attempt at a solution
    So far, to get velocity (s') I took derivative of above equation, giving me v=108-4t^3
    Next, letting v=0 would give me maximum height of particle.
    0=108-4t^3
    t^3=108/-4
    t^3=27
    t=3

    And so at T=3, velocity is 0, and at t<3, velocity is positive.

    Right so far?

    Next step, determine when acceleration is negative. First though, can I reason without using any more math that acceleration is negative when t<3? It seems like it would be, at least in a normal ball projectory question, the ball as it approaches it's max height (in this case t=3) it decreases speed at an increasing rate, which is negative acceleration, and when t>3, it is positive acceleration again.

    In any case, determining acceleration I took derivative of velocity equation (s')
    108-4t^3 ----> -12t^2
    Let acceleration = 0
    0=-12t^2
    12=t^2
    3.464=t

    So when t<3.464, acceleration is negative, and when t>3.464, acceleration is positive.

    My answer for B) doesn't feel right, so I welcome feedback, in fact I really appreciate feedback.

    Regards all
     
  2. jcsd
  3. Mar 18, 2015 #2

    SammyS

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    Right so far.
    Wrong.

    Is t2 ever negative?

    Then is -12 t2 ever positive?

    Check your algebra.
    0 = -12t2 does not give 12 = t2 .

     
  4. Mar 20, 2015 #3
    Would I take -12t^2 and say:
    -12t * (t) = 0

    -12t=0 ---> t=0
    The whole thing equals zero?
     
  5. Mar 20, 2015 #4

    SteamKing

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    When you move a number (108) to the opposite side of the = sign, what must you also do?

    You've got to be careful. Silly mistakes can lead to wrong answers.
     
  6. Mar 20, 2015 #5
    Correct. This works better.
    0=108-4t3
    -108=-4t3
    -108/4=t3
    -3=t
    Thank you.
     
  7. Mar 20, 2015 #6

    SteamKing

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    No, you're still making silly mistakes with your algebra.

    Take the equation:
    108 - 4t3 = 0

    In order to get an expression containing t on one side, we subtract 108 from both sides of the equation:

    108 - 108 - 4t3 = 0 - 108

    which leaves

    -4t3 = -108

    In order to clear the factor of -4 from the LHS, divide both sides of the equation by -4, thus:

    -4t3 / -4 = -108 / -4 or

    t3 = 27

    solving for t by taking the cube root of both sides:

    [t3]1/3 = 271/3

    therefore, t = 3

    In general, unless there are some unusual circumstances, a negative value of t is to be rejected as unrealistic.

    You should refresh yourself on basic algebra and arithmetic.
     
  8. Mar 20, 2015 #7
    Yes I should.
    Thanks again.
     
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