d(t) = (1/3)t^3 + (3/2)t^2 - 4t + c
when t = 1
d(t) = 10.5 + c
This is its displacement after one second, now it turns around and goes back (velocity = 0)
d(t) = (1/3)t^3 + (3/2)t^2 - 4t + c
when t = 3
d(t) = -13/6 + c
this is its displacement when t = 3 (the limit given in the question).
Obviously, when t = 0, d(t) = c
So the particle has started at (c), gone to 10.5 +c, then back to (-13/6) + c, how far has it gone? Well thats 10.5 * 2 + 13/6 = 23.1666667 metres, or (139/6) m .
You have to think logically here, it has gone 10.5 away, then to get to point (-13/6), it must have gone back (38/3) units in the other direction. In this case you are thinking of the distance traveled on the 'round trip' if you like. I hope this helps, the main thing you need to understand in this question is that acceleration, displacement, and velocity can all be found (through integration of derivation) as long as you know one of them. Also, check the answer yourself, I may have made some adding/substracting errors along the way.