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Math Help - Distance traveled

  1. #1
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    Distance traveled

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

    v(t) = 3t - 5 for 0 less than or equal to t less than or equal to 3

    I know how to do part a. The dispalcement is -3/2, but I don't know how to do part b.

    I know that the find the distance traveled you need to integrate the absolute value of v(t), and to do that you must split the integral into 2 parts, one where v(t) is less than or equal to 0 and one where v(t) is greater than or equal to zero.

    Can someone please explain how you split the integrals? I have no problem integrating 3t - 5, but I don't know what intervals to evaluate it at.

    Thanks
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  2. #2
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    Quote Originally Posted by zachb View Post
    The velocity function (in meters per second) is given for a particle moving along a line. Find (a) the displacement and (b) distance traveled.

    v(t) = 3t - 5 for 0 less than or equal to t less than or equal to 3

    I know how to do part a. The dispalcement is -3/2, but I don't know how to do part b.
    The displacement is,
    \int_0^3 3t - 5 dt =-1.5
    The distance is,
    \int_0^3 |3t-5|dt = \frac{20}{3}

    We need to split the integrals for the absolute value.
    Now 3t-5 \geq 0 for t\geq 5/3
    This is where we make the seperation.
    \int_0^{5/3} |3t-5| dt + \int_{5/3}^3 |3t-5|dt
    \int_0^{5/3} - (3t-5)dt + \int_{5/3}^3 (3t-5)dt
    Last edited by ThePerfectHacker; May 21st 2007 at 07:37 PM.
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  3. #3
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    Okay, I understand how you split the integral now, but I don't understand what you did in the last step.

    You started with the definite integral from 0 to 5/3 plus the definite integral from 5/3 to 3, but in the last step you changed it to the definite integral from 0 to 3 plus the definite integral from 5/3 to 3. Why? Is that a mistake, because it doesn't make sense to me.
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  4. #4
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    Quote Originally Posted by zachb View Post
    Okay, I understand how you split the integral now, but I don't understand what you did in the last step.

    You started with the definite integral from 0 to 5/3 plus the definite integral from 5/3 to 3, but in the last step you changed it to the definite integral from 0 to 3 plus the definite integral from 5/3 to 3. Why? Is that a mistake, because it doesn't make sense to me.
    Usually when something does not make sense it is wrong. Judge Judy loves to say that. But to answer your question, yes, I made a mistake and fixed it now.
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