# Thread: velocity and distance with integration

1. ## 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

2. Originally Posted by cowboys111
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 = $\displaystyle \int_{\frac {\pi}2}^{2 \pi} |v(t)|~dt$

3. $\displaystyle \int_{\frac {\pi}2}^{2 \pi} |v(t)|~dt$

$\displaystyle \int_{\frac {\pi}2}^{\frac {3\pi}2} |v(t)|~dt$

4. Originally Posted by cowboys111
$\displaystyle \int_{\frac {\pi}2}^{2 \pi} |v(t)|~dt$

$\displaystyle \int_{\frac {\pi}2}^{\frac {3\pi}2} |v(t)|~dt$
... um, what are you doing here?

5. 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)?

6. Originally Posted by cowboys111
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)?
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

7. ok, so did they figure out where the integral was positive or negative by graphing it?

8. Originally Posted by cowboys111
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

9. awesome. thanks for the help

10. 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