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Math Help - Integrals: Total Distance

  1. #1
    Super Member angel.white's Avatar
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    Integrals: Total Distance

    he velocity function v(t) (in meters per second) is given for a particle moving along a line.

    v(t)=3t-7, 0\leq t\leq 3

    a) Find the displacement d1 traveled by the particle during the time interval given above.

    b) Find the total distance d2 traveled by the particle during the time interval given above.


    -----

    I found a) is -7.5, but I'm a bit confused by integrals. I don't understand how calculating the area under the graph finds "displacement" (I'm a bit confused about what displacement actually means in this kind of a question). And I have no idea how to approach part b (it doesn't even sound like it should involve integrals).

    Also, if anyone has a particularly firm grasp on this, and would be willing to write a brief paragraph explaining why the Fundamental Theorem of Calculus works (not a proof, just an explanation, or a push in the right direction for my own mind), I would really appreciate you for that.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by angel.white View Post
    he velocity function v(t) (in meters per second) is given for a particle moving along a line.

    v(t)=3t-7, 0\leq t\leq 3

    a) Find the displacement d1 traveled by the particle during the time interval given above.

    b) Find the total distance d2 traveled by the particle during the time interval given above.


    -----

    I found a) is -7.5, but I'm a bit confused by integrals. I don't understand how calculating the area under the graph finds "displacement" (I'm a bit confused about what displacement actually means in this kind of a question). And I have no idea how to approach part b (it doesn't even sound like it should involve integrals).
    The displacement of an object in this context pretty much means "position measured from an origin." It is a vector quantity, that is to say it has direction. So a displacement of -7.5 units means that the displacement is a vector 7.5 units long in the negative direction.

    Now, this object started on the positive side of the origin and moved to the negative side. Obviously we need to include the whole trip to find the distance. Mathematically the way we do this is
    d = \int |v(t)| ~dt

    So your d2 will be:
    d2 = \int_0^3 |3x - 7|~dt

    -Dan
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  3. #3
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    Hello, angel.white!

    Try not to worry about the concept of "area".

    We know that the derivative of the position function, s(t),
    . . is the velocity function, v(t).

    Hence, the position function, s(t), is the integral of the velocity, v(t).

    That's all we need to understand.


    The velocity for a particle moving along a line: . v(t)\:=\:3t-7,\;\; 0 \leq t\leq 3

    a) Find the displacement d_1 traveled by the particle during the time interval.
    We have: . s(t) \;=\;\int(3t-7)\,dt \;=\;\frac{3}{2}t^2 - 7t + C

    s(0) \;=\;\frac{3}{7}(0^2) -7(0) + C \;=\;C

    s(3) \;=\;\frac{3}{7}(3^2) - 7(3) + C \;=\;C - \frac{15}{2}

    The displacement is: . d_1\:=\:\left(C - \frac{15}{2}\right) - C \:=\:-\frac{15}{2}


    The particle ends up 7.5 meters to the left of its starting point.



    b) Find the total distance d_2 traveled during the time interval.

    When t=0, the velocity is: , v(0) \:=\:3(0) - 7 \:=\:-7
    . . The particle is moving to the left.

    The velocity is zero when: . 3t-7 \:=\:0\quad\Rightarrow\quad t \:=\:\frac{7}{3} seconds.
    . . The particle stops for an instant.

    After that, the particle is moving to the right.


    When t=0\!:\;s(0) \:=\:C

    When t = \frac{7}{3}\!:\;s\left(\frac{7}{3}\right) \:=\:\frac{3}{2}\left(\frac{7}{3}\right)^2 - 7\left(\frac{7}{3}\right) + C \:=\:C - \frac{49}{6}

    The displacement is: . \left(C - \frac{49}{6}\right) - C \:=\:-\frac{49}{6}
    . . The particle moved \frac{49}{6} meters to the left.


    When t = 3\!:\;s(3)\:=\:C - \frac{15}{2}

    The displacement is: . \left(C - \frac{15}{2}\right) - \left(C - \frac{49}{6}\right) \:=\:\frac{2}{3}
    . . The particle moved \frac{2}{3} meters to the right.


    Therefore: . d_2 \:=\:\frac{49}{6} + \frac{2}{3} \:=\:\frac{53}{6}\:=\:8\frac{5}{6} meters.

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