Recovering Motion from Initial Data

Integrating acceleration gives velocity; integrating velocity gives position. These are two separate integrations, so determine a separate constant each time. The condition may be given at any time, not only at the start.

If acceleration is \(a(t)\) and \(v(t_0)=v_0\), then \(v(t)=v_0+\int_{t_0}^{t}a(u)\,\mathrm du\). If \(s(t_0)=s_0\), then \(s(t)=s_0+\int_{t_0}^{t}v(u)\,\mathrm du\). The letter \(u\) is only the integration variable.

For the illustrative constant acceleration \(a=2\), with \(v(1)=3\) and \(s(1)=-4\), integration gives \(v=2t+C_1\), hence \(C_1=1\). Then \(s=t^2+t+C_2\), hence \(C_2=-6\). Substitution at \(t=1\) checks both conditions.

At a change of motion model, carry the position and velocity at the end of the first stage into the second stage. If the second stage uses a new clock \(T=t-t_0\), its initial conditions occur at \(T=0\). A new formula does not imply a return to the origin or to rest.