Maths ( Quadratic Equations )

Let x be the speed of the first train
y be the speed of the second train
t be the total time for the first train travel from A to B
u be the total time for the second train travel from A to B
Using speed x time = distance

We obtain two equations, xt=540---------(1)
2y+(y+5)(u-2)=540----------(2)

When we consider the time that the two trains travel, we can obtain the equation
t-1=u-------(3) (because they arrive at same time, but first train start an hour early)

Also, we can consider the point the second train reach when the first train has passed 40 mins earlier

2y=(3-40/60)x (The distance from A to the above time is the same)
=2 1/3x
y=7/6x---------(4)

As we can see, there are 4 unknowns in the above equations, hence we need to find 4 equations to solve them

Put (3) into (2)
2y+(y+5)(t-3)=540-------(5)

Put (1) into (5)
2y+(y+5)(540/x-3)=540-------(6)

Put (4) into (6)
2y+(y+5)[540(7)/6y-3]=540
2y(6y)+(y+5)[3780-18y]=540(6y)
12y^2+3780y-18y^2+18900-90y=3240y where ^2 is to the power of 2 (square)
6y^2-450y-18900=0
y^2-75y-3150=0

y=[75+{75^2-4(-3150)}]/2 or y=[75-{75^2-4(-3150)}]/2 {}is square root

=[75+{18225}]/2 or =[75-{18225}]/2

=[75+135]/2 or =[75-135]/2
=105 or =-30 (rej as it must be positive) (negative means opposite position which is impossible in this case)

Put y=105 into (4)
x=90

Hence, the speed of the first train when it left A is 90km/h
the speed of the second train when it left A is 105km/h

Let x km/hr be the speed of the first train.
Let y km/hr be the speed of the second train.

Time needed for the first train = 540/x hours.
Time needed for the second train = (540/x - 1) hours.

At the passed point,
Time needed for the second train = 2 hours.
Time needed for the first train
= 2 hours - 40 mins + 1 hour
= 2 hours 20 mins
= (2 + 1/3) hours
= 7/3 hours

Distanced travelled at that point = 7/3 x km = 2y km
7x = 6y
y/x = 7/6

For the remaining journey,
Time needed for the second train
= (540/x - 1 - 2) hours.
= (540/x - 3) hours.

Distanced travelled for the remaining journey
= (540/x - 3) * (y + 5) km
= (540y/x - 3y + 540*5/x - 15) km
= (630 - 3y + 2700 * 7/6/y - 15) km
= (3150/y + 615 - 3y) km

Distanced travelled for the entired journey
= (3150/y + 615 - 3y) km + 2y km
= (3150/y + 615 - y) km
3150/y + 615 - y = 540
3150/y + 75 - y = 0
y - 75 - 3150/y = 0 and y GT 0
y^2 - 75y - 3150 = 0 and y GT 0
(y + 30) (y - 105) = 0 and (y + 30) GT 30
y - 105 = 0
y = 105
x = 6/7 y
= 90

GT = is greater than

The speed of the first train is 90 km/hr.
The speed of the second train is 105 km/hr for the first 2 hours and then increases to 110 km/hr.