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Math Help - Placement And Displacement

  1. #1
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    Placement And Displacement

    Hi my friends ,, again
    Sorry for asking to many question in short time but what can i do , i am studying calculus all by myself ( wish me luck )
    in the following question

    what the different when it comes to Placement and Displacement
    should i divide the integration when i try find the displacement ??
    i dont need u to solve the question for me , i just want to understand the general rule for this case

    Thanks in advanced
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  2. #2
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    "Placement" isn't normally encountered in this sort of a problem. I suppose it might mean "starting somewhere". In the absence of other information, one might assume that one starts at the Origin.

    Displacement is simply how far one has travelled.

    Always, and the notation is ubiquitous, is this relationship for constant acceleration.

    s(t) = -Żat^{2} + v_{0}t + h_{0}

    h_{0} is the place to start - your "placement", I suppose. If you start at zero, s(0) = 0 = h_{0}.

    This leaves: s(t) = -Żat^{2} + v_{0}t

    The first derivative: s'(t) = v(t) = at + v_{0} gives the velocity.

    The second derivative: s"(t) = v'(t) = a(t) = a gives the acceletation.

    It is this last equation that will fill your head with solutions.

    s"(t) = v'(t) = a(t)

    Given v(t), find v'(t) to get the acceleration.
    Given v(t), find the antiderivative to get the displacement. You will also need to know where to start.

    Givenm then v(t) = t^2 - t, we have s(t) = (1/3)t^3 - (1/2)t^2 + h_{0}. In this case, since we are asked only how far things move, not where they land, we can ignore the initial displacement so that h_{0} = 0.

    Simple enough, then, s(5) = (1/3)125 - (1/2)25 = 25((1/3)5 - (1/2))

    That's enough of that. You tell me why Part B is even a different question. They look rather similar, don't they?

    Hint: Is our friendly particla ALWAYS moving in the same direction?
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  3. #3
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    Hello, metallica007!

    Who uses the term "placement"?
    The location of the particle is usually called its Position.


    A particle moves along a line with the velocity function: . v(t) \:=\:t^2-t
    . . where v is measured in meters per second.

    Find:

    (a) the displacement during the time interval [0, 5]
    The position function is the integral of the velocity function.

    s(t) \;=\;\int (t^2 - t)\,dt \;=\;\tfrac{1}{3}t^3 - \tfrac{1}{2}t^2 + s_o

    . . where s_o is the particle's initial position (at t = 0).


    At t=0\!:\;\;s(0) \:=\:\tfrac{1}{3}(0^3) - \tfrac{1}{2}(0^2) + s_o \;=\;s_o

    At t = 5\!:\;\;s(5) \;=\;\tfrac{1}{3}(5^3) - \tfrac{1}{2}(5^2) + s_o \;=\;\frac{175}{6} + s_o


    During that time interval, its displacement is:
    . . s(5) - s(0) \;=\;\left(\frac{175}{6}+s_o\right) - s_o \;=\;\frac{175}{6} \;=\;29\tfrac{1}{6}\text{ meters}



    (b) the distance traveled during the time interval [0, 5].
    The particle changes direction when v = 0.

    . . t^2-t \:=\:0 \quad\Rightarrow\quad t(t-1) \:=\:0 \quad\Rightarrow\quad t \;=\;0,\;1


    At t=0, its position is: . s(0) = s_o

    At t = 1, its position is: . s(1) \:=\:\tfrac{1}{3}(1^3) - \tfrac{1}{2}(1^2) + s_o \;=\;s_o-\tfrac{1}{6}

    Its displacement is: . s(1) - s(0) \;=\;\left(s_o - \tfrac{1}{6}\right) - s_o \;=\;-\tfrac{1}{6}
    . . In the first second, the particle moved \tfrac{1}{6} meter to the left.


    At t = 5, its position is: . s(5) \;=\;\tfrac{1}{3}(5^3) - \tfrac{1}{2}(5^2) + s_o \;=\;\tfrac{175}{6} +s_o

    Its displacement is: . s(5) - s(1) \;=\;\left(\tfrac{175}{6} + s_o\right) - \left(s_o - \tfrac{1}{6}\right) \;=\;\tfrac{88}{3}
    . . In the next 4 seconds, it moved 29\tfrac{1}{3} meters to the right.


    Therefore, it moved a total distance of: . \tfrac{1}{6} + 29\tfrac{1}{3} \;=\;29\tfrac{1}{2} meters.

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, metallica007!

    Who uses the term "placement"?
    The location of the particle is usually called its Position.


    The position function is the integral of the velocity function.

    s(t) \;=\;\int (t^2 - t)\,dt \;=\;\tfrac{1}{3}t^3 - \tfrac{1}{2}t^2 + s_o

    . . where s_o is the particle's initial position (at t = 0).


    At t=0\!:\;\;s(0) \:=\:\tfrac{1}{3}(0^3) - \tfrac{1}{2}(0^2) + s_o \;=\;s_o

    At t = 5\!:\;\;s(5) \;=\;\tfrac{1}{3}(5^3) - \tfrac{1}{2}(5^2) + s_o \;=\;\frac{175}{6} + s_o


    During that time interval, its displacement is:
    . . s(5) - s(0) \;=\;\left(\frac{175}{6}+s_o\right) - s_o \;=\;\frac{175}{6} \;=\;29\tfrac{1}{6}\text{ meters}



    The particle changes direction when v = 0.

    . . t^2-t \:=\:0 \quad\Rightarrow\quad t(t-1) \:=\:0 \quad\Rightarrow\quad t \;=\;0,\;1


    At t=0, its position is: . s(0) = s_o

    At t = 1, its position is: . s(1) \:=\:\tfrac{1}{3}(1^3) - \tfrac{1}{2}(1^2) + s_o \;=\;s_o-\tfrac{1}{6}

    Its displacement is: . s(1) - s(0) \;=\;\left(s_o - \tfrac{1}{6}\right) - s_o \;=\;-\tfrac{1}{6}
    . . In the first second, the particle moved \tfrac{1}{6} meter to the left.


    At t = 5, its position is: . s(5) \;=\;\tfrac{1}{3}(5^3) - \tfrac{1}{2}(5^2) + s_o \;=\;\tfrac{175}{6} +s_o

    Its displacement is: . s(5) - s(1) \;=\;\left(\tfrac{175}{6} + s_o\right) - \left(s_o - \tfrac{1}{6}\right) \;=\;\tfrac{88}{3}
    . . In the next 4 seconds, it moved 29\tfrac{1}{3} meters to the right.


    Therefore, it moved a total distance of: . \tfrac{1}{6} + 29\tfrac{1}{3} \;=\;29\tfrac{1}{2} meters.

    Thank u i get it now
    if i integrated from 0 to 5 i get the total displacement and if i divide the integration i get the total distance
    distance>=displacement

    Thank you
    Last edited by metallica007; January 8th 2010 at 10:06 PM.
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