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Math Help - Magnitue

  1. #1
    Junior Member R3ap3r's Avatar
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    Magnitue

    I was doing just fine on my homework until I came across this bad boy. The x and y coordinates of a moving particle are given by the parametric equations x = t^3 = 5t and y = 4t^2 - 3t^3 where t is in seconds, and x and y are in metres. To 3 significant digits, find the magnitute and direction of the acceleration at t=1\; s.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by R3ap3r View Post
    I was doing just fine on my homework until I came across this bad boy. The x and y coordinates of a moving particle are given by the parametric equations x = t^3 = 5t and y = 4t^2 - 3t^3 where t is in seconds, and x and y are in metres. To 3 significant digits, find the magnitute and direction of the acceleration at t=1\; s.
    r(t)=(t^3+5t) \vec i + (4t^2-3t^3) \vec j

    taking the derivative

    v(t)=(3t^2+5) \vec i + (8t-9t^2) \vec j

    one more time

    a(t)=(6t) \vec i + (8-18t) \vec j

    The direction is given by a(1)

     a(1)=6 \vec i - 10 \vec j

    the magnitue is

     |a(1)|=\sqrt{(6)^2+(-10)^2}=\sqrt{136}=2\sqrt{34}
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  3. #3
    Junior Member R3ap3r's Avatar
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    Alright so the answer is 11.7 magnitue. How do I find the direction?
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by R3ap3r View Post
    Alright so the answer is 11.7 magnitue. How do I find the direction?
    a(1)=6 \vec i - 10 \vec j

    this is the vector pointing in the direction of acceleration.

    if you need the angle we take the arctan

     \theta=\tan^{-1}\left( \frac{y}{x}\right)=\tan^{-2}\left( \frac{-10}{6}\right) \approx -59.04^\circ
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