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Math Help - velocity and distance with integration

  1. #1
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    velocity and distance with integration

    Hi, im given a velocity function v(t)=cos(t); pi/2 <= t <= 2pi

    I am able to find the displacement but not the distance, does anyone know how to do this?
    thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cowboys111 View Post
    Hi, im given a velocity function v(t)=cos(t); pi/2 <= t <= 2pi

    I am able to find the displacement but not the distance, does anyone know how to do this?
    thanks
    Distance = \int_{\frac {\pi}2}^{2 \pi} |v(t)|~dt
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  3. #3
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     <br />
\int_{\frac {\pi}2}^{2 \pi} |v(t)|~dt<br />

    \int_{\frac {\pi}2}^{\frac {3\pi}2} |v(t)|~dt
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cowboys111 View Post
     <br />
\int_{\frac {\pi}2}^{2 \pi} |v(t)|~dt<br />

    \int_{\frac {\pi}2}^{\frac {3\pi}2} |v(t)|~dt
    ... um, what are you doing here?
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  5. #5
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    sorry about the last post. How does the book get from \int_{\frac {\pi}2}^{2 \pi} |v(t)|~dt

    to a negative integral going from pi/2 to 3pi/2 cot(t) + integral going from 3pi/2 to 2pi cos(t)?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cowboys111 View Post
    sorry about the last post. How does the book get from \int_{\frac {\pi}2}^{2 \pi} |v(t)|~dt

    to a negative integral going from pi/2 to 3pi/2 cot(t) + integral going from 3pi/2 to 2pi cos(t)?
    because |x| = \left \{ \begin{array}{lr}  x & \mbox { if } x \ge 0 \\ & \\ -x & \mbox{ if } x < 0\end{array} \right.

    so they split the integral into two parts. the interval on which it is negative and the interval on which it is positive. on the interval the function was negative, they took the negative of the function (which is positive) for that interval and integrated it
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  7. #7
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    ok, so did they figure out where the integral was positive or negative by graphing it?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cowboys111 View Post
    ok, so did they figure out where the integral was positive or negative by graphing it?
    yes, that's one way. another way is to solve for the function = 0, then test points in all the intervals to see if you get a negative or positive answer
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  9. #9
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    awesome. thanks for the help
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  10. #10
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    Okay, think I got it.

    I split it up:
    cosine is negative from \pi/2 to 3\pi/2 so for that duration, I changed it to -cosine

    \int_{\pi/2}^{3\pi/2} -cos(t) ~dt + \int_{3\pi/2}^{2\pi} cos(t) ~dt

    which turned into:
     -sin(3\pi /2) + sin( \pi /2) + sin(2 \pi)-sin(3 \pi /2)

    which equals 3
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