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Vectors - Displacement to Accelleration

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data
    An object moves with its position obeying x=5t^2-6t^2 and y=3t^-1+4t^4. What is the magnitude of its acceleration at t=1s;
    a)57 m/s^2
    b)34 m/s^2
    c)26 m/s^2
    d)44 m/s^2
    e)0 m/s^2


    2. Relevant equations
    x=5t^3-6t^2
    y=3t^-1+4t^4

    3. The attempt at a solution
    Second derivative of the displacement x is;
    ax=30t-12 m/s^2
    Second derivative of the displacement y is;
    ay=8t m/s^2

    t=1 -> ax
    ax = 18

    t=1 -> ay
    ay = 8

    Resultant acceleration = sqrt(18^2+8^2) = 19.69

    Which is wrong according to the answer I found online which was a) 57 m/s^2. Anyone show me where I've messed up or is the answer I found wrong?
     
    Last edited: Nov 17, 2011
  2. jcsd
  3. Nov 17, 2011 #2

    PeterO

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    Homework Helper

    The x expressions don't match up, but I guess you actually meant x=5t^3-6t^2

    I can't imagine what you thought the y expression was ??
     
  4. Nov 17, 2011 #3
    Ahh yes sorry about that, I'll fix that up. What do you mean by 'I can't imagine what you thought the y expression was?' That equation for y is the exact one given?
     
  5. Nov 17, 2011 #4
    Can you check that

    x=5t^2-6t^2 and y=3t^-1+4t^4

    are correct.
     
  6. Nov 17, 2011 #5
    x=5t3-6t2
    y=3t-1+4t4
    Are the the correct equations given in the question.
     
  7. Nov 17, 2011 #6

    PeterO

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    Homework Helper

    OK .. if that is the expression for y, then 8t is not the second derivative.
     
  8. Nov 17, 2011 #7
    Ahh I got the answer this time, my problem was I thought I could simplify y=3t-1+4t4 to 4t3/3 I now realize what a rookie error that was lol. Thanks for your help everyone.
     
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