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
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
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