Question 44544
<pre><font size = 4><b>A car leaves a town at 40 miles per hour. Two hours later
a second car leaves the town traveling the same route and 
overtakes the first car in 5 hours and 20 minutes. how 
fast was the second car traveling?

Let the rate of the second car be x mph

Make this chart:

            D      R     T
1st car    
2nd car  

We can fill in the rats of the cars as 40 mph and x mph

            D      R     T
1st car           40   
2nd car            x   

We can fill in the time of the 2nd car as 5 hours 20 minutes,
that is, after we change this to 5 1/3 hours and then to an 
improper fraction 16/3

            D      R     T
1st car           40       
2nd car            x   16/3

The 1st car left 2 hours before the 2nd car, so we add 2 hours
to the 1st car's 5 hours 20 minutes and get 7 hours 20 minutes 
or 7 1/3 hours or as an improper fraction 22/3 hours.

            D      R     T
1st car           40   22/3    
2nd car            x   16/3

Now we use D = RT to find the distances:

For first car:  D = RT = 40(22/3) = 880/3
For second car: D = RT = (x)(16/3) = (16/3)x

            D      R     T
1st car   880/3   40   22/3
2nd car  (16/3)x   x   16/3

Now we use the fact that both cars traveled
the same distance to equate the two distances:

880/3 = (16/3)x

Solve that and get x = 55 mph

Edwin
AnlytcPhil@aol.com</pre>