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

Printable View

- Dec 4th 2007, 03:34 PMcowboys111velocity 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 - Dec 4th 2007, 03:35 PMJhevon
- Dec 4th 2007, 03:44 PMcowboys111
$\displaystyle

\int_{\frac {\pi}2}^{2 \pi} |v(t)|~dt

$

$\displaystyle \int_{\frac {\pi}2}^{\frac {3\pi}2} |v(t)|~dt$ - Dec 4th 2007, 03:47 PMJhevon
- Dec 4th 2007, 03:48 PMcowboys111
sorry about the last post. How does the book get from $\displaystyle \int_{\frac {\pi}2}^{2 \pi} |v(t)|~dt$

to a negative integral going from pi/2 to 3pi/2 cot(t) + integral going from 3pi/2 to 2pi cos(t)? - Dec 4th 2007, 03:50 PMJhevon
because $\displaystyle |x| = \left \{ \begin{array}{lr} x & \mbox { if } x \ge 0 \\ & \\ -x & \mbox{ if } x < 0\end{array} \right.$

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 - Dec 4th 2007, 04:06 PMcowboys111
ok, so did they figure out where the integral was positive or negative by graphing it?

- Dec 4th 2007, 04:07 PMJhevon
- Dec 4th 2007, 04:08 PMcowboys111
awesome. thanks for the help

- Dec 4th 2007, 04:15 PMangel.white
Okay, think I got it.

I split it up:

cosine is negative from $\displaystyle \pi/2$ to $\displaystyle 3\pi/2$ so for that duration, I changed it to -cosine

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

which turned into:

$\displaystyle -sin(3\pi /2) + sin( \pi /2) + sin(2 \pi)-sin(3 \pi /2)$

which equals 3