SOLUTION: 2 CYCLIST START BIKING FROM A TRAIL START 3 HOURS APART. THE 2ND TRAVELS AT 10 MILES PER HOUR AND STARTS 3 HOURS AFTER THE 1ST WHO IS TRAVELING AT 6 MILES PER HOUR, HOW MUCH TIME

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Question 273582: 2 CYCLIST START BIKING FROM A TRAIL START 3 HOURS APART. THE 2ND TRAVELS AT 10 MILES PER HOUR AND STARTS 3 HOURS AFTER THE 1ST WHO IS TRAVELING AT 6 MILES PER HOUR, HOW MUCH TIME WILL PASS BEFORE SECOND CATCHEDS FIRST
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Each cyclist must have their own d=r%2At equation
1st cyclist:
d%5B1%5D+=+r%5B1%5D%2At%5B1%5D
2nd cyclist:
d%5B2%5D+=+r%5B2%5D%2At%5B2%5D
given:
The 1st cyclist has a head start which is
6 mi/hr x 3 hrs = 18 mi
r%5B1%5D+=+6 mi/hr
r%5B2%5D+=+10 mi/hr
-----------------------
Assume I have a stopwatch and I start it when
the 2nd cyclist leaves. When I stop the watch
t%5B1%5D+=+t%5B2%5D so I'll call them both t
The 2nd cyclist has to make up the head start, so
d%5B2%5D+=+d%5B1%5D+%2B+18 mi
------------------------
d%5B1%5D+=+r%5B1%5D%2At
(1) d%5B1%5D+=+6t
and
d%5B2%5D+=+r%5B2%5D%2At
(2) d%5B1%5D+%2B+18+=+10t
I'll substitute (1) into (2)
6t+%2B+18+=+10t
4t+=+18
t+=+4.5 hrs
This is the elapsed time after the 1st cyclist got
a 3 hr head start, so
the 2nd cyclist catches the 1st cyclist 4.5+%2B+3+=+7.5 hrs
after the 1st cyclist left
check answer:
d%5B2%5D+=+10t
d%5B2%5D+=+10%2A4.5
d%5B2%5D+=+45 mi
and
d%5B1%5D+=+6t+
and, since
d%5B2%5D+=+d%5B1%5D+%2B+18
d%5B1%5D+=+d%5B2%5D+-+18
then,
d%5B2%5D+-+18+=+6%2A4.5
d%5B2%5D+-+18+=+27
d%5B2%5D+=+45 mi
OK