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Simple Acceleration Question

  1. May 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Velocity in m/s at time t is defined as v= 20t(1+2t)^-2

    Find Acceleration at 0.5s

    2. The attempt at a solution

    I missed the day we covered this topic in class and I am under the assumption that I would find the derivative of 20t(1+2t)^-2 and then find the derivative once more to get acceleration. Once I get the second derivative its as easy as substituting 0.5 for t. Unfortunately I never get the correct answer which I was told is 9.6m/s. Is the answer wrong? I end up with 15m/s all the time.
  2. jcsd
  3. May 14, 2007 #2


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    Homework Helper

    If you are given the velocity function you only need to differentiate once to get the acceleration.
  4. May 14, 2007 #3
    I must be doing something wrong I keep getting the wrong answers.


    s(t) = (4t^2 + 5)^3 t=1

    I do the work

    First Derivative - 24t(4t^2 + 5)^2
    Second Derivative - 384t^2(4t^2 + 5)

    My answer is 3456 m/s and the book tells me it is 5400 m/s
  5. May 14, 2007 #4


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    So is this supposed to be a new question? What happened with the first one you posted?

    Note the unit of acceleration is m/s^2.

    Your second derivative is not right. You're using the product rule, right? You're missing a term.
  6. May 14, 2007 #5
    :rofl: I feel really stupid right now. I knew something was up, turns out I was forgetting the product rule for all the questions. Thanks for clueing me into that. What a silly mistake.
  7. May 14, 2007 #6


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    You're welcome, lol. Those things happen. :smile:
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