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Math Help - Kinematics (Calculus Form)

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    Kinematics (Calculus Form)

    Question:
    A particle travels along a straight line which passes through A and B. During the motion the velocity of the particle is 6t+t^2 metres per second, where t seconds is the time measured from a certain instant. The particle passes through A when t=2 and through B when t=5. Find, in terms of t, the acceleration of the particle and its distance from A at any instant during the motion. Calculate, also, the distance of B from A.


    My Answers:
     a =  \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}(6t+t^2)}{\mathrm{d}t} = 6 + 2t

    v = (6t+t^2) = \frac{\mathrm{d}x}{\mathrm{d}t}
    \therefore x = 3t^2 + \frac13 t^3 + c
    x_{\vec{AB}} = \left[3t^2 + \frac13 t^3\right]^5_2 = 102


    My Problem:
    I don't know how to do this part (don't even understand what it is saying ): 'Find...its distance from A at any instant during the motion'*. Can someone help? Thanks in advance.

    *The answer is x = 3t^2+ \frac13 t^3 - \frac{44}{3}.
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  2. #2
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    Quote Originally Posted by Air View Post
    Question:
    A particle travels along a straight line which passes through A and B. During the motion the velocity of the particle is 6t+t^2 metres per second, where t seconds is the time measured from a certain instant. The particle passes through A when t=2 and through B when t=5. Find, in terms of t, the acceleration of the particle and its distance from A at any instant during the motion. Calculate, also, the distance of B from A.


    My Answers:
     a =  \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}(6t+t^2)}{\mathrm{d}t} = 6 + 2t

    v = (6t+t^2) = \frac{\mathrm{d}x}{\mathrm{d}t}
    \therefore x = 3t^2 + \frac13 t^3 + c
    x_{\vec{AB}} = \left[3t^2 + \frac13 t^3\right]^5_2 = 102


    My Problem:
    I don't know how to do this part (don't even understand what it is saying ): 'Find...its distance from A at any instant during the motion'*. Can someone help? Thanks in advance.

    *The answer is x = 3t^2+ \frac13 t^3 - \frac{44}{3}.
    The particle is at A when t = 2. So
    x_A = 3(2)^2 + \frac13 (2)^3 + c = \frac{44}{3} + c

    Thus the particle's distance from A will be
    x - x_A = \left ( 3t^2 + \frac13 t^3 + c \right ) - \left ( \frac{44}{3} + c \right ) = 3t^2 + \frac{1}{3}t^3 - \frac{44}{3}

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    The particle is at A when t = 2. So
    x_A = 3(2)^2 + \frac13 (2)^3 + c = \frac{44}{3} + c

    Thus the particle's distance from A will be
    x - x_A = \left ( 3t^2 + \frac13 t^3 + c \right ) - \left ( \frac{44}{3} + c \right ) = 3t^2 + \frac{1}{3}t^3 - \frac{44}{3}

    -Dan
    How did you know that it is x - x_A and not x_A - x?
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  4. #4
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    Quote Originally Posted by Air View Post
    How did you know that it is x - x_A and not x_A - x?
    When t > 2, \, x > x_A and so clearly the distance is x - x_A. HOWEVER .......

    When 0 < t < 2, \, x < x_A and so the distance will be x_A - x ...... It's apparent that the question actually wanted the distance from at any instant during the motion for t > 2.
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