# Displacement and Distance

#### alane1994

##### Active member
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!

~Austin

#### Jameson

Staff member
Displacement is the integral of the velocity vector over the given time interval and results in the distance between the starting point and end point. If you move 10 meters north and 10 meters south, the displacement is 0.

To find the distance you need to figure out where the displacement is negative and count that as a positive value. I believe this is the same as taking the integral of the absolute value of velocity. The way I would do this is find the regions where v(t) is positive and negative and then breaking the calculation into multiple calculations.

When considering $$\displaystyle v(t)=15\cos{3t}$$ where $$\displaystyle 0 \le t \le 2\pi$$ at $t=0$ v(t) is positive and becomes negative when $$\displaystyle 3t=\frac{\pi}{2}$$ so when $$\displaystyle t=\frac{\pi}{6}$$.

So your first integral is $$\displaystyle \int_{0}^{\frac{\pi}{6}}v(t)dt$$ This value will be positive.

Now v(t) will be negative until until it touches the x-axis again where $$\displaystyle 3t=\frac{3\pi}{2}$$ or when $$\displaystyle t=\frac{\pi}{2}$$. The second integral is now $$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}v(t)dt$$. This will be negative but when considering distance you don't need to take into account the sign of this so count it as positive.

If you keep repeating this over until the end of the interval you should get the final answer. It will take a while to calculate this way and there very well could be a quicker way to do it but that's how I would do this.

#### alane1994

##### Active member
Yeah, that's what I did after consulting a professor, I got the answer 60...

#### Jameson

Staff member
Yeah, that's what I did after consulting a professor, I got the answer 60...
You already found all of these intervals for part A so this calculation shouldn't have been too tedious. Wow your professor responds fast. You posted this question just a couple of hours ago

#### topsquark

##### Well-known member
MHB Math Helper
Just to put my two cents in...

The formula for distance is
$$dist = \int |v|dt$$

where v is the velocity vector.

In a practical sense this can pretty much only be calculated in the way Jameson has described.

-Dan