# 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?

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