
Bus A with speed Sa and bus B with speed Sb. Then in the first hypothesis we will have (B starts 2 hour after A):
Sa(t) + Sb(t2)=220 and
Sa(t)=110 (1) and
Sb(t2)=110 (2)
So t=110/Sa=110/Sb + 2
So 110/Sa=110/Sb + 2 or multiplying by Sa*Sb:
110Sb=110Sa+2SaSb or
(Sa55)(Sb+55)+55^2=0 (3)
In the 2nd hypothesis:
Sa(4) + Sb(4)=220 so
Sa+Sb= 55. (4)
From (3) and (4)
Sb(Sb+55)=55^2 or
Sb^2+55Sb55^2=0 or
Sb =(55+sqrt(5*55^2))/2 = (55+55sqrt(5))/2=34 km/h approx.
and Sa=55Sb=5534=21 km/h approx.
This is a classical opposite direction problem. The basic formula is:
Speed1*t + speed2(tt0) =Distance apart.
t0 being the time of inactivity since mobile 1 started. 