1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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

    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.

    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

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Right so far.

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
    Thank you.
  7. Mar 20, 2015 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted