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Math Help - HELP!!!! NEED HELP FINDING MAXIMUM VELOCITY...

  1. #1
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    Angry HELP!!!! NEED HELP FINDING MAXIMUM VELOCITY...

    how do you find the maximum velocity of

    s(t)=-93.75t^3+93.75t^2+45t

    step by step to the answer would help a ton...
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  2. #2
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     s(t)=-93.75t^3+93.75t^2+45t
    The derivative of position, s(t), gives you velocity

    v(t)=-281.25t^2+187.5t+45

    Now, do you know how to find the maximum of a function?
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  3. #3
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    Quote Originally Posted by Linnus View Post
     s(t)=-93.75t^3+93.75t^2+45t
    The derivative of position, s(t), gives you velocity

    v(t)=-281.25t^2+187.5t+45

    Now, do you know how to find the maximum of a function?
    can you do it step by step to the answer so i can see how to do it right

    thanks
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  4. #4
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    Assuming that you know how to use the power rule...

     s(t)=-93.75t^3+93.75t^2+45t
    The derivative of position, s(t), gives you velocity

    v(t)=-281.25t^2+187.5t+45

    The derivative of velocity is...
    a(t)=-562.5t+187.5
    set it equal to 0
     a(t)=-562.5t+187.5=0
    I hope you can solve t...and get t=\frac {1}{3}

    Use the first derivative test to see if it is a maximum (it is)
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  5. #5
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    Quote Originally Posted by Linnus View Post
    Assuming that you know how to use the power rule...

     s(t)=-93.75t^3+93.75t^2+45t
    The derivative of position, s(t), gives you velocity

    v(t)=-281.25t^2+187.5t+45

    The derivative of velocity is...
    a(t)=-562.5t+187.5
    set it equal to 0
     a(t)=-562.5t+187.5=0
    I hope you can solve t...and get t=\frac {1}{3}

    Use the first derivative test to see if it is a maximum (it is)
    so after -562.5t+187.5=0
    its 187.5 X -3 = 562.5t

    can you show me how the max and min is found leading to the max velocity please
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  6. #6
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    ....
    a(t)=-562.5t+187.5=0<br />
    Solve for t..
    -562.5t=-187.5

    t=\frac{-187.5}{-562.5}=\frac {1}{3}

    I surely hope you knew how to do that.
    Also, that is WHERE the maximum velocity occurred at on v(t).
    To find the maximum velocity, you would need to plug in 1/3 into V(t).

    To determine whether it is a max or a min, you need to do a first derivative test in this case. I cannot teach you how to do it, that is something your teacher should do.
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  7. #7
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    so you just plug in v(t) like this to find max velocity

    -281.25(1/3)^2+187.5t(1/3)^2+45 = ?
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  8. #8
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    yes...
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  9. #9
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    yes...except...where did you get the "t(1/3)^2" in

    -281.25(1/3)^2+187.5t(1/3)^2+45 = ?

    because that wasn't in the v(t) equation...
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