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Thread: Velocity, Accelertion problem

  1. #1
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    Velocity, Accelertion problem

    Let s(t)=2t28t+2 for 0t7 denote the position of an object moving along a line. Find

    a) Velocity at time t: v(t)=s'(t)= 4t-8

    b) Acceleration at time t: v'(t)= 4

    c) Initial position: 2(0)-8(0)+2= 2

    d) Ending position: 2(7)^2-8(7)+2= 44

    --I am having trouble with this part--
    e) Total distance traveled: I went the simple route since it started at 2 and ended at 44, I answered 42, but that's not correct.

    f) Velocity is positive on the interval: (2,7]

    --And this part--
    g) Acceleration is positive on the interval [0,??
    Since it was a closed interval for the equation, and the acceleration shows to be constant, 4, I tried [0,7], and [0,7), neither of which is right.
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  2. #2
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    Re: Velocity, Accelertion problem

    Quote Originally Posted by JamieG View Post
    Let s(t)=2t28t+2 for 0t7 denote the position of an object moving along a line. Find
    a) Velocity at time t: v(t)=s'(t)= 4t-8
    b) Acceleration at time t: v'(t)= 4
    c) Initial position: 2(0)-8(0)+2= 2
    d) Ending position: 2(7)^2-8(7)+2= 44
    --I am having trouble with this part--
    e) Total distance traveled: I went the simple route since it started at 2 and ended at 44, I answered 42, but that's not correct.
    f) Velocity is positive on the interval: (2,7]
    --And this part--
    g) Acceleration is positive on the interval [0,??
    Since it was a closed interval for the equation, and the acceleration shows to be constant, 4, I tried [0,7], and [0,7), neither of which is right.
    Note that the velocity $v'(t)<0,~0\le t<2$ So $s(0)-s(2)=2-(-6)=8$
    Then the particle goes $s(7)-s(2)=~?$

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