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.
Let Vx be the velocity of X and Vy be the velocity of Y, and t=0 be the time that X left Powersdale.
At t=3 Y has traveled 2*Vy.
45 minutes before was t=2.25, by which time x had gone 2.25*Vx, so
2*Vy = 2.25*Vx
To avoid messy fractions define v = Vy/9 = Vx/8.
After time t=3, Y must go an additional 240  18*v km, traveling at (9*v+5) km/h, thus taking (24018*v)/(9*v+5) hours. During this same time, X must travel whatever the remainder is of its 240 km trip, so that is 240/(8*v)  3. Thus we have
(24018*v)/(9*v+5) = 240/(8*v)  3
Simplifying and crossmultiplying, we get a quadratic
3v^2  5*v  50 = 0
which solves to v = 5/6 +/ 25/6.
We don't want a negative velocity, so v = 30/6 = 5.
So Vx = 8*v = 40 km/h and Vy = 9*v = 45 km/h
As a check, at t=2.25 X is at the 90 km mark
At t=3 Y is also at the 90 km mark
At t=6 X is at 240, which is Quicksville
and Y is at 90+50*3 = 240 (Quicksville).

Posted by Charlie
on 20070320 09:51:39 