# Total Distance Travelled vs. Change in Position

There's a difference between how far someone travels, and the change between their starting and ending positions. Take a basketball player running suicide drills. He travels to one line and back to his starting point, then to a farther line and back to his starting point, and does that several times. By the time he's done, he's certainly travelled some distance. However, since he ends up back where he started, his change in position is 0.

When we integrate a velocity function from *t* = *a* to *t* = *b*, the number we get is the change in position between *t* = *a* and *t* = *b*. If we want to know the total distance travelled, we have to do something different. We don't want travel in opposite directions canceling out, so we find how far we travel in one direction

and how far we travel in the opposite direction

and then, instead of subtracting, we add these distances together. We're pretending all the distances travelled are in the same direction. Another way to think of this is that we're integrating the absolute value of the velocity function, |*v*(*t*)|.

This gets us the total (unweighted) area between the function *v*(*t*) and the horizontal axis.

### Sample Problem

A cat climbs a tree with velocity given by the graph below.

If the cat starts at ground level (height *s*(0) = 0 feet),

(a) how high is the cat when *t* = 8 seconds?

(b) what is the total distance the cat travels from *t* = 0 to *t* = 8?

Answers.

(a) This part of the question is like ones we did earlier. We want to know the cat's change in position from *t* = 0 to *t* = 8, so we integrate the velocity function by looking at the areas on the graph. We find that

so the cat's position at *t* = 8 is *s*(8) = 12 feet.

(b) This part of the question is asking for the total distance the cat travelled, which means we want to find .

To do this we count all the areas between the graph of *v*(*t*) and the horizontal axis positively, to get

This means the cat traveled a total distance of 18 feet.