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Math Help - can someone help with this please

  1. #1
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    can someone help with this please

    The angular displacement, qradians, of the spoke of a wheel is given by
    the expression q= 0.5t4 t3, where t is time in seconds. Find:

    (a) The angular velocity after 2 seconds

    (b) The angular acceleration after 3 seconds


    (c) The time when the angular acceleration is zero
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  2. #2
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    Re: can someone help with this please

    Quote Originally Posted by tomjay View Post
    The angular displacement, qradians, of the spoke of a wheel is given by
    the expression q= 0.5t4 t3, where t is time in seconds. Find:

    (a) The angular velocity after 2 seconds

    (b) The angular acceleration after 3 seconds


    (c) The time when the angular acceleration is zero
    Velocity is the derivative of displacement, and acceleration is the derivative of velocity.
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  3. #3
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    Re: can someone help with this please

    a)differentiate the expression once and substitute t=2.
    b)differentiate the expression twice and substitute t=3
    c) Equate the twice differentiated expression to zero and find 't'.
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  4. #4
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    Re: can someone help with this please

    What ive done
    a) Velocity = d Ɵ /dt

    =2t^3 3t^2 = V

    at t = 2,

    velocity = 2(2^3) 3(2^2)
    =16 12
    = 4 Radians/second

    b) Acceleration = dv/dt

    = 6t^2 6t =A

    At t = 3

    Acceleration = 6 (3^2) 6 (3)
    = 54 18
    36 Radians/second^2


    c) A = dv/dt = 6t^2 6t = 0

    6t^2 = 6t

    Acceleration = 0 when t = 1 second
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  5. #5
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    Re: can someone help with this please

    correct...Did you ask a doubt or r u testing our knowledge?
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  6. #6
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    Re: can someone help with this please

    Quote Originally Posted by tomjay View Post
    What ive done
    a) Velocity = d Ɵ /dt

    =2t^3 3t^2 = V

    at t = 2,

    velocity = 2(2^3) 3(2^2)
    =16 12
    = 4 Radians/second

    b) Acceleration = dv/dt

    = 6t^2 6t =A

    At t = 3

    Acceleration = 6 (3^2) 6 (3)
    = 54 18
    36 Radians/second^2


    c) A = dv/dt = 6t^2 6t = 0

    6t^2 = 6t

    Acceleration = 0 when t = 1 second
    Always solve quadratics by factorising or using the Quadratic Formula/Completing the Square. By dividing both sides by t, you are assuming that \displaystyle \begin{align*} t \neq 0 \end{align*}.

    \displaystyle \begin{align*} 6t^2 - 6t &= 0 \\ 6t(t - 1) &= 0 \\ 6t = 0 \textrm{ or } t- 1 &= 0 \\ t = 0 \textrm{ or } t &= 1 \end{align*}

    There is also 0 acceleration when t = 0.
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  7. #7
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    Re: can someone help with this please

    Quote Originally Posted by tomjay View Post
    What ive done
    a) Velocity = d Ɵ /dt

    =2t^3 3t^2 = V

    at t = 2,

    velocity = 2(2^3) 3(2^2)
    =16 12
    = 4 Radians/second
    Yes, that is correct.

    b) Acceleration = dv/dt

    = 6t^2 6t =A

    At t = 3

    Acceleration = 6 (3^2) 6 (3)
    = 54 18
    36 Radians/second^2
    Yes, that is correct.

    c) A = dv/dt = 6t^2 6t = 0

    6t^2 = 6t

    Acceleration = 0 when t = 1 second
    This is a quadratic equation and has two solutions. Yes, one solution is t= 1. What is the other?
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