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Math Help - Rectilinear Motion: finding displacement & distance traveled w/ integration (2nd ? )

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    Rectilinear Motion: finding displacement & distance traveled w/ integration (2nd ? )

    v(t) = t^3 - 3t^2 +2t 0 < t <3

    Alright, I was fine finding the displacement ...

    I did sigma(t^3-3t^2 + 2t)dt

    I got t^3-3t^2 + 2t and evaluated on 0-3

    (81/4 - 108/4 + 36/4) = 9/4 meters


    However I am having trouble finding the total distance traveled....


    Since it is over the period of 0-3, I know I need to split up the integrals when the sign of v(t) changes |(t^3-3t^2 + 2t)dt|

    Setting v(t) to 0 I get 0, 1, 2 ... The sign changes from negative to positive at t=1

    I took sigma 0,1 |t^3-3t^2 + 2t dt|- sigma 1,2 |t^3-3t^2 + 2t dt|

    and got some funky answer

    Not sure what my mistake was... please help...I have a test on this tomorrow and I am so lost
    The solutions manual says the answer is 11/4 meters
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Rectilinear Motion: finding displacement & distance traveled w/ integration (2nd

    The velocity function may be factored as follows:

    v(t)=t(t-1)(t-2)

    So, we see there are roots at t=0,\,1,\,2. Testing the sign we find v(t)<0 on (1,2) and so the total distance traveled is:

    D=\int_0^1 t^3-3t^2+2t\,dt-\int_1^2 t^3-3t^2+2t\,dt+\int_2^3 t^3-3t^2+2t\,dt

    Can you proceed?
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    Re: Rectilinear Motion: finding displacement & distance traveled w/ integration (2nd

    I will try when I get around to it today and post back but it looks good to me

    I think that's what I did, I might have made some arithmetic mistake but I'll just have to try it again and post what I got if it's wrong ....

    Thanks
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