Rates of Change, particle velocity/distance?

[shocklightnin]

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

ive set t = 17 but it doesn't 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

 

[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.

 

[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?

 

[LCKurtz]

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?

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