SOLUTION: A JOGGER STARTS FROM ONE END OF A 15 MILE NATURE TRAIL AT 8:00 AM ONE HOUR LATER A CYCLIST STARTS FROM THE OTHER END OF THE TRAIL AND RIDES TOWARD THE JOGGER IF THE RATE OF THE JOG

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A JOGGER STARTS FROM ONE END OF A 15 MILE NATURE TRAIL AT 8:00 AM ONE HOUR LATER A CYCLIST STARTS FROM THE OTHER END OF THE TRAIL AND RIDES TOWARD THE JOGGER IF THE RATE OF THE JOG      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 720607: A JOGGER STARTS FROM ONE END OF A 15 MILE NATURE TRAIL AT 8:00 AM ONE HOUR LATER A CYCLIST STARTS FROM THE OTHER END OF THE TRAIL AND RIDES TOWARD THE JOGGER IF THE RATE OF THE JOGGER IS 6MPH AND THE RATE OF THE CYCLIST IS 9MPH AT WHAT TIME WILL THE TWO MEET?
THANK YOU
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
You can think of the jogger as standing still and
the cyclist approaching at the sum of their speeds
-----------
The jogger cuts down the distance between them by
+d%5B1%5D+=+6%2A1+
+d%5B1%5D+=+6+ mi
So now the cyclist only has to travel +15+-+6+=+9+ mi
-----------
Let +t+ = the time in hrs it takes the cyclist to meet the jogger
+9+=+%28+6+%2B+9+%29%2At+
+9+=+15t+
+3+=+5t+
+t+=+3%2F5+ hrs
+%283%2F5%29%2A60+=+36+ min
The two will meet at 8:36 AM
-----------
check:
jogger's equation:
+d%5B2%5D+=+6%2A%283%2F5%29+
+d%5B2%5D+=+18%2F5+ mi
Cyclist's equation:
+d%5B3%5D+=+9%2A%283%2F5%29+
+d%5B3%5D+=+27%2F5+
+d%5B2%5D+%2B+d%5B3%5D+ must = 9 mi
+18%2F5+%2B+27%2F5+=+45%2F5+
+45%2F5+=+9+
OK