SOLUTION: Russ and Janet are running in the Apple Hill Fun Run. Russ runs at 7 mph, Janet runs at 5 mph. If they start at the same time, how long will it be before they are half a mile apart

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Russ and Janet are running in the Apple Hill Fun Run. Russ runs at 7 mph, Janet runs at 5 mph. If they start at the same time, how long will it be before they are half a mile apart      Log On


   

Question 177643: Russ and Janet are running in the Apple Hill Fun Run. Russ runs at 7 mph, Janet runs at 5 mph. If they start at the same time, how long will it be before they are half a mile apart? show me step by step..
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Russ and Janet are running in the Apple Hill Fun Run. Russ runs at 7 mph, Janet runs at 5 mph. If they start at the same time, how long will it be before they are half a mile apart? show me step by step..

You can do it in your head. But your teacher won't accept that. But I'll show you how to do it in your head anyway, then we'll do it by algebra.
In your head:
Their rate of separation is 7-5 or 2 miles an hour. So in 1 hour they would be 2 miles apart. In half an hour they would be 1 mile apart, so in half that time, or 15 minutes, they will be half a mile apart. Answer: 15 minutes.
But, of course, your teacher won't accept that. So,
By algebra:
Let t = the time in hours they both run before Russ is
ahead by 1%2F2 mile.


Make this DRT-chart

         Distance      Rate        Time
Russ         
Janet     

They both ran for t hours. So fill in t for both
their times:


         Distance      Rate        Time
Russ                                 t  
Janet                                t

Now fill in 7 for Russ' rate and 5 for Janet's rate:


         Distance      Rate        Time
Russ                     7           t  
Janet                    5           t

Now use formula DISTANCE = RATE × TIME
t fill in their distances:

         Distance      Rate        Time
Russ       7t            7           t  
Janet      5t            5           t

Now the difference between their distances 
must equal 1%2F2 mile:

Russ' distance MINUS Janet's distance = 1%2F2 a mile

So:

              7t - 5t = 1%2F2
 
                   2t = 1%2F2

Multiply both sides by 2 to clear of fractions:

                   4t = 1

Divide both sides by 4

                    t = 1%2F4

That's 1%2F4 an hour or 15 minutes.

Edwin