# Rates of Change, particle velocity/distance?

1. Mar 8, 2012

### shocklightnin

1. The problem statement, all variables and given/known data

The position of a particle is given by the equation
s = 2t^3 + 36t^2 − 168t
where t is measured in seconds and s in meters.

What is the total distance traveled by the particle after the first seventeen seconds?

2. Relevant equations

s(t) = 2t^3 + 36t^2 − 168t
s'(t) = 6t^2+72t-168 = v(t)

3. The attempt at a solution

ive set t = 17 but it doesnt give me the right answer. (i get 17374). what am i doing wrong?! i know that the particle is at rest after 2 seconds (by taking derivative of this function to get the velocity function). the particle is moving forward in time interval t>2 and moving backward in time interval t<2.R

2. Mar 8, 2012

### LCKurtz

If you calculated s(17) - s(0), that gives the change in position between t = 0 and t = 17. But that isn't the total distance traveled unless the particle never reverses direction. If it changes direction you have to include the distance going to and fro too.

3. Mar 9, 2012

### shocklightnin

so calculating the direction is from 0-2 and 2-17? so it would be:
176+176+17550? (for total distance without any signs to indicate direction)
does that mean the particle went backwards 176, then forwards 176, then forwards 17550 from point 2 to 17?

4. Mar 9, 2012

### LCKurtz

That's the right calculation except you might re-check you calculation for $t=17$.