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Traveling Trains and Nonlinear Systems

Date: 07/16/98 at 08:05:56
From: Jack Shen
Subject: Solving linear-quadradic system algebraically

Town A1 and B1 are 540 km apart. A train leaves A1 and travels to B1 at 
a uniform speed. A second train leaves A1 one hour later and after two 
hours it reaches a point passed by the first train 40 minutes earlier. 
The second train then increases its speed by 5 km/h and arrives at B1 
at the same time as the first train. What were the speeds of the trains 
when they left A1? 

I have no idea how to start this question. 

Date: 07/24/98 at 10:27:06
From: Doctor Erra
Subject: Re: Solving linear-quadradic system algebraically

Hello Jack,

Let us find the equations of your problem. I will call:

   v1  the (uniform) speed of the first train
   v2  the initial speed of the second train
    t  the time used by the first train to reached B1
   x1  the distance between A1 and "the point passed by the first train 
       40 minutes earlier."
   t1  the time used by the first train to reached x1

We use then the classical law x = v*t (distance equals velocity times 
time), to find:

   (1)  540 = v1*t      because speed v1 is uniform
   (2) v2*2 = v1*2/3    because 40 mins = 2/3 h and hypothesis about x1
   (3)  540 = v2*2+(v2+5)*(t-2)   because second train reaches B1 

Let us see this with a "picture":

Train 1:

Train 2:
            540 =  v2*2             +   (v2+5)*(t-2)
     |     v2*2 = v1*2/3           |  

You can see that we have "forgotten" the unknown x1, but we have a 
"strong" system because of the nonlinearity of the equation (3). But 
you can solve it; I've tried it. It's always a good idea to try the 
substitution method. I'll explain this method and let you do the 

First use equation (1). This gives:

   t = 540/v1    (1)'

We can then substitute the new expression of t into equation (3). 
This gives an equation, which I will call (3)', with only v1 and v2. 
In essence, we have "lost" t, and this is good. We now use equation 
(2), rewriting it as:
   v2 = v1/3     (2)' 

Now, again, we substitute this new expression of v2 into equation (3)', 
giving us equation (4).

When you simplify the left and right sides of equation (4), you will 
see that we have a nonlinear equation with only one unknown, the speed: 
v1. To solve the problem, it suffices to solve equation (4), because if 
you known v1, you can find v2 and t, by (1)' and (2)'.

To verify your own results, the value of t is 74. I gave you a lot of 
computations to do but I'm sure you will be able to do them. If you 
have some trouble solving equation (4), please write back.

- Doctor Erra, The Math Forum
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