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Math Help - Velocity, Displacement, Distance

  1. #1
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    Velocity, Displacement, Distance

    The velocity function is v(t)= -t^2 + 6t -8 for a particle moving along a line. Find the displacement and the distance travelled by the particle during the time interval [-1, 5].
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  2. #2
    nvv
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    Hi, My Little Pony

    displacement = integral from -1 to 5 (-t^2 + 6t -8) = -18

    distance = Sqrt[ ((-5^2+6*5-8) - (1^2-6*1-8))^2) ] = Sqrt[((-3)-(-12))^2]=Sqrt[12^2] = 12 (I'm not sure about this but it looks correctly)

    p.s. I'm new on this forum and I don't now how to use math symbols , can anyone help me ?
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  3. #3
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    Hello, My Little Pony!

    The velocity function is: v(t)\:= \:-t^2 + 6t -8 for a particle moving along a line.
    Find the displacement and the distance travelled on the time interval [-1, 5].
    We have: . v(t) \:=\:-t^2 + 6t - 8

    The position function is: . s(t) \:=\:-\tfrac{1}{3}t^3 + 3t^2 - 8t + C

    Assume that s(0) = 0, then: C = 0\quad\Rightarrow\quad s(t) \:=\:-\tfrac{1}{3}t^3 + 3t^2 - 8t


    s(\text{-}1) \:=\:-\tfrac{1}{3}(\text{-}1)^3 + 3(\text{-}1)^2 - 8(\text{-}1) \:=\:\frac{34}{3}

    s(5) \:=\:\tfrac{1}{3}(5^3) + 3(5^2) - 8(5) \:=\:-\frac{20}{3}

    The displacement is: . \left(-\frac{20}{3}\right) - \left(\frac{34}{3}\right) \:=\:-\frac{54}{3} \:=\:-18

    The particle is displaced 18 units to the left.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Solve v(t) = 0

    t^2 - 6t + 8 \:=\:0 \quad\Rightarrow\quad (t-2)(t-4) \:=\:0 \quad\Rightarrow\quad t \:=\:2,\:4

    . . . This is where the particle stops ... and changes directions.


    s(\text{-}1) \:=\:-\tfrac{1}{3}(\text{-}1) + 3(1)  -8(\text{-}1) \:=\:\frac{34}{3}
    s(2) \:=\:-\tfrac{1}{3}(8) + 3(2^2) - 8(2) \:=\:-\frac{20}{3}
    In the first two seconds, its displacement is: . -\frac{20}{3} - \frac{34}{3} \:=\:-18
    It has moved 18 units to the left.


    s(2) \:=\:-\frac{20}{3}
    s(4) \:=\:-\tfrac{1}{3}(64) + 3(16) - 8(4) \:=\:-\frac{16}{3}
    In the next two seconds, its displacement is: . -\frac{16}{3} -\left(-\frac{20}{3}\right) \:=\:\frac{4}{3}
    It has moved \frac{4}{3} units to the right.


    s(4) = -\frac{16}{3}
    s(5) \:=\:-\frac{20}{3}
    In the last second, its displacement is: . \left(-\frac{20}{3}\right) - \left(-\frac{16}{3}\right) \:=\:-\frac{4}{3}
    It has moved \frac{4}{3} units to the left.


    It has moved a total distance of: . 18 + \frac{4}{3} + \frac{4}{3} \:=\:\frac{62}{3} \:=\:20\tfrac{2}{3} units.

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