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Math Help - Integral

  1. #1
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    Integral

    The velocity function (in meters per second) is given for a particle moving along a line.
    a. Find displacement
    b. Distance traveled.

    V(t)= 3t -5 , 0 <= t <= 3
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    The velocity function (in meters per second) is given for a particle moving along a line.
    a. Find displacement
    b. Distance traveled.

    V(t)= 3t -5 , 0 <= t <= 3
    Here
    Attached Thumbnails Attached Thumbnails Integral-dist1.gif  
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    Here
    answer for
    a. -3/2 m
    b. 41/6 m
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    answer for
    a. -3/2 m
    b. 41/6 m
    ok, so apparently i mixed up the questions, i got the answer for (a)

    the answer was negative since the part of the curve was under the x-axis (or t-axis) was larger than the part above it. for distance travelled however, we cannot have a negative value. so we must find these two areas in terms of magnitude and add them together. to find the areas as positive maginitudes i needed to split the integral in 2 and turn the limits of the first backward. that gives me positive areas for both pieces
    Attached Thumbnails Attached Thumbnails Integral-distance.gif  
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    is up to his old tricks again! Jhevon's Avatar
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    Here's the graph so you see what i'm talking about:
    Attached Thumbnails Attached Thumbnails Integral-graph.jpg  
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  6. #6
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    Hello, camherokid!

    the velocity function (in m/sec) is given for a particle moving along a line.
    . . v(t) .= .3t - 5, .0 < t < 3

    (a) Find displacement
    (b) Distance traveled
    The position function is: .s(t) .= .∫(3t - 5) dt .= .3t/2 - 5t + C


    (a) When t = 0: .s(0) .= .30/2 - 50 + C .= .C

    . . .When t = 3: .s(3) .= .33/2 - 53 + C .= .C - 3/2

    The displacement is: .(C - 3/2) - C .= .-3/2

    (The particle is 1 units to the left of where it started.)


    (b) The velocity is zero when t = 5/3
    . . .(That's when the particle changes direction.)

    We know that: .s(0) = C

    When t = 5/3: .s(5/3) .= .C - 25/6
    . . The displacement is: .(C - 25/6) - C .= .-25/6
    It has moved 25/6 units to the left.

    When t = 3: .s(3) .= . C - 3/2
    . . The displacement is: .(C - 3/2) - (C - 25/6) .= .8/3
    It has moved 8/3 units to the right.

    Therefore, the total distance is: .25/6 + 8/3 .= .41/6 units.

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