Four cars travel at constant speed along a road. Three of them (A, B and C) travel in one direction, and the fourth, D, in the opposite direction.
Car A passes B and C at 9:00h and 10:00h, respectively, and crosses car D at 11:00h.
Car D crosses car B and car C at 13:00h and 15:00h, respectively.
At what time did car B pass car C?
Bonus point: can you solve this geometrically?
Consider the speeds as relative to car D (i.e., taking car D to be the "stationary" point of reference). Refer to the speeds as Sa, Sb and Sc.
Car C takes 5 hours from its being with car A to its passing car D, while car A takes only one hour from that event, so
Sa = Sc * 5
Car B takes 4 hours from its being with car A to its passing car D, while car A takes only two hours, so
Sa = Sb * 2
Sb = Sc * 5/2
The positions of cars C and B at a given time are given by
Pc = Sc * (15 - t) as C meets D at 15 hours.
Pb = Sb * (13 - t) as B meets D at 13 hours.
We need the time that Pb = Pc, so that Pb/Pc = 1, so we can divide one equation by the other:
1 = (5/2) * (13-t)/(15-t)
t = 11 + 2/3, or 11:40.
Posted by Charlie
on 2008-06-09 13:40:19