Distance Traveled

Well, to get the 'big picture', you can make a sketch.

$12t^2 - 36t + 15 = 3(4t^2 - 12t + 5) = 3(2t - 5)(2t - 1)$

(it's strange that it's very similar to http://www.mathhelpforum.com/math-he...tml#post605939)

You'll see that the graph a large part of the graph is negative between 0.5 and 2.

This means that the particle what initially going forward, but decelerating until changing direction. Then in accelerated, but within the 2nd second, it's still moving backwards. The integration that you gave does not give the distance, but the displacement, that is the final position of the particle, relative to the starting point.

To find the distance covered, you'll need to work out:

$\displaystyle \int^{0.5}_0 12t^2 - 36t + 15\ dt + \int^2_{0.5} -(12t^2 - 36t + 15)\ dt$