SOLUTION: 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

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: 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       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 44544: 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?
Found 2 solutions by Nate, AnlytcPhil:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
Since the first car was given a two-hour head start, it traveled 2(40 mi/h) = 80 miles ....
First Car's Rate: 40t + 80
Second Car's Rate: xt
The different of the two car's traveled distance is equal to zero:
40t + 80 - xt = 0
40(16/3) + 80 - (16/3)x = 0
-(16/3)x = -240/3 - 640/3
-(16/3)x = -880/3
x = (880/3)(3/16) = 55
The second car was traveling 55 miles per hour.
Check:
First Car: y = 40x + 80
Second Car: y = 55x
Where 'y' is the distance and 'x' is the time:
-y + 40x = -80
+
y - 55x = 0
-15x = -80
x = 16/3
5 hours and 20 minutes

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
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