Question

How do you find the total displacement for the particle whose position at time `t` is given by `s(t)=3t^2-5t` on the interval `0<=t<=2`?

Answer 1

The total displacement will be the difference between the final and initial positions.
You have:
Initial position; for `t=0`
`s_i=s(0)=3×0^2-5×0=0`
Final position; for `t=2`
`s_f)=s(2)=3×(2)^2-5×2=12-10=2`

Total displacement`=s_f)-s_i=2-0=2`
(Km, m, cm, mm....)