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Math Help - Kinematics

  1. #1
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    Kinematics

    Question:
    A particle moves along a straight line starts at time t=0 seconds with a velocity 4 \ ms^{-1}. At any subsequent time t seconds the acceleration of the particle is (6t-8) \ ms^{-2}. Find:
    (a) The distance moves before first coming to instantaneous rest,
    (b) the total time T seconds taken by the particle to return to the starting point,
    (c) the greatest speed of the particle for 0 \le t \le T.


    My Answers (Without Method):
    (a) x = \frac{32}{27} \ m
    (b) T = 2 \ s


    My Problem in this question:
    I don't understand part (c). Here is my method:

     v = 3t^2 - 8t + 4
     \therefore Greatest speed when \frac{\mathrm{d}v}{\mathrm{d}t} = 0. (At turning point?)
    \therefore \frac{\mathrm{d}v}{\mathrm{d}t} = 6t-8=0 \implies t = \frac43
    \therefore v = 3\left(\frac43\right)^2 - 8\left(\frac43\right) + 4 = -\frac43
    \text{Speed} = |v| = \frac43 ?!!!

    How would part (c) be done? The correct answer is 4 \ ms^{-1}. Thanks in advance.
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  2. #2
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    Quote Originally Posted by Air View Post
    Question:
    A particle moves along a straight line starts at time t=0 seconds with a velocity 4 \ ms^{-1}. At any subsequent time t seconds the acceleration of the particle is (6t-8) \ ms^{-2}. Find:
    (a) The distance moves before first coming to instantaneous rest,
    (b) the total time T seconds taken by the particle to return to the starting point,
    (c) the greatest speed of the particle for 0 \le t \le T.


    My Answers (Without Method):
    (a) x = \frac{32}{27} \ m
    (b) T = 2 \ s


    My Problem in this question:
    I don't understand part (c). Here is my method:

     v = 3t^2 - 8t + 4
     \therefore Greatest speed when \frac{\mathrm{d}v}{\mathrm{d}t} = 0. (At turning point?)
    \therefore \frac{\mathrm{d}v}{\mathrm{d}t} = 6t-8=0 \implies t = \frac43
    \therefore v = 3\left(\frac43\right)^2 - 8\left(\frac43\right) + 4 = -\frac43
    \text{Speed} = |v| = \frac43 ?!!!

    How would part (c) be done? The correct answer is 4 \ ms^{-1}. Thanks in advance.

    Your asked to find the maximum speed in the given interval 0-2 seconds.

    as v = 3t^2 - 8t + 4 the maximum speed is unbounded with time increasing, your calculus method found the minimum speed of the particle. Draw a rough sketch of the velocity function and the answer will be obvious.

    Bobak
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