SOLUTION: A train leaves station 1 at 7:00am for station 2, which is 1080 kilometers away. At 10:00am, a faster train leaves station 2 for station 1 and runs 15kph faster than the first trai

Question 476507: A train leaves station 1 at 7:00am for station 2, which is 1080 kilometers away. At 10:00am, a faster train leaves station 2 for station 1 and runs 15kph faster than the first train. When do the trains pass each other if this occurs midway between station 1 and 2?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Train 1 7.00 AM --- x kph
Train 2 10.00 AM--- x+15 kph
Distance between two stations = 1080
Midway = 540 km
AT 10.00 AM Train A is 540-3x from A
Distance between them at 10.00AM = 1080-3x
combined speed = x+x+15=2x+15 kph

From 10:00 AM the time taken for them to meet is the same.
Time taken by them to meet = Time taken by Train B to meet
(1080-3x)/(2x+15)= 540/(x+15)
(x+15)(1080-3x)=540(2x+15)
1080x-3x^2+16200 -45x= 1080x+8100
-3x^2-45x+8100=0
3x^2+45x-8100=0
Find the roots of the equation by quadratic formula
a= -3 ,b= -45 ,c= 8100
b^2-4ac=
b^2-4ac= 2025 - 97200
b^2-4ac= 99225
= 315

x1=( 45 + 315 )/ -6
x1= -60
x2=( 45 -315 ) / -6
x2= 45
Train 1 speed = 45 kph which is positive
Train A speed = 45 kph
Train B speed = 60 kph
..
Time taken by train B to meet = 540/60 = 9 hours
Train B starts at 10:00 AM
They meet at 10 + 9 = 19:00 hours OR 7:00 PM
m.ananth@hotmail.ca
m.ananth@hotmail.ca

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