Recovering position from velocity

If v=ds/dt, then s= v,dt. Position requires an integration constant. Why: differentiating s+C gives the same velocity for every constant C. A known position selects one journey from these parallel position graphs. Method: integrate with respect to time; insert the given position and its time to find C; then solve the appropriate position equation in the allowed domain. For v=2t-2 and s(0)=3, s=t2-2t+3. Returning to the initial point means s=3, giving t=0,2; the next return is at 2 s. Check: a negative velocity does not make the position negative. Returning to the origin means s=0; returning to the initial point means s=s(0).