Train X moves at an uniform speed from Powersdale to Quicksville; two stations 240 kilometers apart. Train Y starts from Powersdale precisely one hour after Train X departed (also from Powersdale) and, after two hours, comes to a point that Train X had passed 45 minutes previously.
Train Y now increases its speed by 5 kilometers per hour and it overtakes Train X as soon it reaches Quicksville.
Determine the original speed of both the trains.
(In reply to
solution by Charlie)
Call the total time T so the velocity of train X is 240/t
Since train Y can make up 1/4 of an hour on train X in only 2 hours after starting 1 hour behind, its velicity is 9/8 of train X or 270/t
So we have train Y covering 240 miles in two legs. The first takes 2 hours at (270/t) kph. The second takes (t-3) hours at (270/t + 5) kph.
240 = 2(270/t) + (t-3)(270/t + 5)
240 = 5t + 255 - 270/t
5t + 15 - 270/t = 0
5tē + 5t - 270 = 0 is quadratic with solutions -9 and 6.
t=6 hours gives velocities of 40 and 50 kph
|
Posted by Jer
on 2007-03-21 11:58:01 |