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Re: time and distance...
by Said Saqid - Wednesday, 13 February 2013, 06:17 AM
  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(t-2)=220 and
Sa(t)=110 (1) and
Sb(t-2)=110 (2)
So t=110/Sa=110/Sb + 2
So 110/Sa=110/Sb + 2 or multiplying by Sa*Sb:
110Sb=110Sa+2SaSb or
(Sa-55)(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+55Sb-55^2=0 or
Sb =(-55+sqrt(5*55^2))/2 = (-55+55sqrt(5))/2=34 km/h approx.
and Sa=55-Sb=55-34=21 km/h approx.

This is a classical opposite direction problem. The basic formula is:
Speed1*t + speed2(t-t0) =Distance apart.
t0 being the time of inactivity since mobile 1 started.