A relation between the law of multiplication with rate of travel and time. I am reading the book Algebra for the practical man, chapter 1, article 6. 
Uptil article-6 the author has explained the operations of addition on positive and negative numbers. Now in article-6 he is explaining(not defining) the multiplication using the concept of negative time.   

  
*
  
*In the 4th point the author presumes that the multiplication of rate with time will give the distance moved.
  

In the light of this concept he concludes that $−ve×−ve=+ve $. 
 The author states in the beginning of article-6:   

We shall consider motion to the right of the starting point as positive ($+$) and motion to the left of the starting point as negative ($-$).

Then in the  4th point he says:

If the train is now at the starting point and has been traveling to the left, where was it 5 hours ago?
  Motion to the left $\times$ Past time
  $(- 40)      \times  (-5) = +200 $  

Here ($-40$) is the rate at which the train moves and ($-5$) is time elapsed.
My question is:

  
*
  
*As the time changes from $0$ to $-5$ the train moves from the starting point towards right in the past, shouldn't we assume the rate of train positive and conclude$(+40)      \times  (-5) = +200 $?.
  
*If my interpretation is correct then does it mean that the author's assumption that
rate $\times$ time = distance is incorrect?
  

NOTE: Since division has not been defined uptil article 6 we cannot use the definition of division.
 A: The rate is negative because the train is moving toward the left.
The time is negative because it's five hours before the time we're calling zero.
Negative rate times negative time equals positive distance, and positive means it's to the right of the location we're calling zero.
You say as time changes from $0$ to $-5$, the train moves rightward.  Rightward does indeed correspond to positive distance.  The change in time is $-5$ and the change in distance is $+200$ therefore the rate is
$$
\text{rate} = \frac{\text{change in distance}}{\text{change in time}} = \frac{+200}{-5} = -40.
$$
The rate is negative.
A: I don't understand the downvotes to this question either.
The author defines $\mbox{distance}= \mbox{rate} \times \mbox{time}$. Somewhere in his definition of rate, he states that rate is positive if moving right, negative if moving left. Perhaps what wasn't clear is that this definition also assumes that the arrow of time has been fixed: that time is increasing. Therefore, rate is actually independent of (the sign of) time. In other words, when speaking of a rate, the convention should be "with respect to increasing time." You probably know from other places that velocity is distance/time, where time is increasing (positive). The reason for all this is to establish a convention so that when talking about a car moving at 50mph, we both assume that this is defined with respect to increasing time (otherwise we always have to establish a convention, which is very annoying).  Notice that the author does not need the notion of division in any of this. Again, to be precise in his defition, the author needs to mention that the rate (velocity) is with-respect-to increasing time.
Given that definition, the problem should be clear. If we reverse time, then the distance covered correctly reverses as well, so if $d=vt$, then $-d=v(-t)$. 
To clarify even further, some time ago mathematicians establish that $5$ is "positive five" and $-5$ is "negative five." This is a set convention, so it's unlikely that you would think $5$ is negative. In the same way, the concept of rate is fixed with respect to increasing time. The author extends the definition of distance covered to negative time by multiplying by negative time instead, in the same way that you multiply $5$ by $(-1)$ to get $-5$. 
