Incoming: a * (v1+v2) = v2t ... [1]
Catch up later: b*(v2-v 1)=v2t.
So there is: a(v 1+v2)=b(v2-v 1).
(a+b)v 1=(b-a)v2
Namely: v 1=(b-a)v2/(a+b)
Substitute the above formula into [1] to get: t.
a[(b-a)v2/(a+b)+v2]=v2t
t = a[(b-a)/(a+b)+ 1]= 2ab/(a+b)