Displacement and Distance

• October 30th 2012, 11:21 AM
alane1994
Displacement and Distance
Here is my question.

The function $v(t)=15\cos{3t}$, $0 \leq t \leq 2 \pi$, is the velocity in m/sec of a particle moving along the x-axis. Complete parts (a) through (c).

a.Graph the velocity function over the given interval. then determine when the motion is in the positive direction and when it is in the negative direction.
(I have done this part)
b. Find the displacement over the given interval.
(I have done this part)
c. Find the distance traveled over the given interval.
(This is the part that I am fuzzy on)

For this, does it involve previously found information? Or is it a separate set of calculations all its own?
Any help is appreciated!
• October 30th 2012, 11:34 AM
HallsofIvy
Re: Displacement and Distance
The distance function is the integral of the velocity function. The "displacement" is the value of the distance function at $t= 2\pi$. You say you have already done that. And you have found that the velocity function is negative for $\pi< t< 2\pi$ so you need to change the sign for "distance traveled". If yu drive 100 miles east and then 100 miles west, you will be right back where you started. Your "displacement" will be 0 but you have traveled 200 miles.