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Math Help - Displacement and Distance

  1. #1
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    Displacement and Distance

    A particle moves with acceleration a(t)m/sec^2 along the s-axis and has velocity Vm/sec at time t=0. Find the displacement and distance traveled by the particle during the given time interval.

    a(t)=-2;V=3;[1,4]


    I got the answer 9m for both displacement and distance...is this correct?
    Last edited by holly123; January 20th 2009 at 02:09 PM.
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  2. #2
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    Hello, holly123!

    Sorry, both your answers are wrong.


    A particle moves with acceleration: a(t) \,=\,2 along the s-axis
    . . and has velocity v = 3 when t=0.

    Find the displacement and distance traveled during the interval [-4,1]

    We have: . a(t) \:=\:2

    Integrate: . v(t) \:=\:2t + C_1

    When t=0,\:v = 3 \;\hdots \text{so we have: }\:3 \:=\:2(0) + C_1 \quad\Rightarrow\quad C_1 = 3

    . . Hence: . v(t) \:=\:2t+3

    Integrate: . (t) \:=\:t^2 + 3t + C



    When t = \text{-}4\!:\;\;s(\text{-}4) \:=\:(\text{-}4)^2 + 3(\text{-}4) + C \:=\: C + 4
    When t=1\!:\;\;s(1) \:=\:1^2 + 3(1) + C \:=\:C+4

    At the beginning of the interval, the particle is at C + 4.
    At the end of the interval, the particle is still at C + 4.

    . . \boxed{\text{The displacement is 0 (zero).}}



    Set the velocity equal to zero: . 2t + 3 \:=\:0 \quad\Rightarrow\quad t \:=\:-\tfrac{3}{2}
    This is when the particle stops and changes direction.

    When t = \text{-}4\!:\;\;s(\text{-}4) \:=\:C+4
    When t = \text{-}\tfrac{3}{2}\!:\;\;s\left(\text{-}\tfrac{3}{2}\right) \:=\:C - \tfrac{9}{4}
    On the interval \left[\text{-}4,\,\text{-}\tfrac{3}{2}\right], it moved: . \left(C - \tfrac{9}{4}\right) - \left(C + 4\right) \:=\:-\tfrac{25}{4}
    . . It moved \tfrac{25}{4} units to the left.

    When t = \text{-}\tfrac{3}{2}\!:\;\;s\left(\text{-}\tfrac{3}{2}\right) \:=\:C - \tfrac{9}{4}
    When t = 1\!:\;\;s(1) \:=\:C + 4
    On the interval \left[\text{-}\tfrac{3}{2},\:1\right], it moved: . (C+4) - \left(C - \tfrac{98}{4}\right) \:=\:\tfrac{25}{4}
    . . It moved \tfrac{25}{4} units to the right.

    Its total distance traveled is: . \frac{25}{4} + \frac{25}{4} \:=\:\boxed{\frac{25}{2}} units.

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  3. #3
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    a(t)= -2 though....do negatives not matter for acceleration?
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