SOLUTION: Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initially 5.0 km east of the flagpole and is running with

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initially 5.0 km east of the flagpole and is running with       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 65618: Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initially 5.0 km east of the flagpole and is running with a constant velocity of 8.0 km/h due west. What will be the distance of the two runners from the flagpole when their paths cross? ( It is not necessary to convert your answer from kilometers to meters for this problem. You may leave it in kilometers.)
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initially 5.0 km east of the flagpole and is running with a constant velocity of 8.0 km/h due west. What will be the distance of the two runners from the flagpole when their paths cross?
--------------------
At the start runner A and runner B are 11 km apart.
---------
Eastbound runner DATA:
distance=x km ; rate= 9 k/h ; time= d/r=x/9 hr
--------
Westbound runner DATA:
distance = 11-x ; rate= 8 k/h ; time = d/r = (11-x)/8
-----------
EQUATION:
time east= time west
x/9= (11-x)/8
8x=99-9x
17x=99
x=5.92 km (distance eastbound runner goes)
He will meet the westbound runner 6-5.92=0.08 km west of the flagpole.
Cheers,
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The runner running west has velocity v%5B1%5Dkm/hr
He runs adistance d%5B1%5Dkm
and meets the other runner after t%5B1%5D hours
----------------------------
The runner running east has velocity v%5B2%5Dkm/hr
He runs adistance d%5B2%5Dkm
and meets the other runner after t%5B2%5D hours
-----------------------------
Their velocities and therefore, distances are different,
but they each run for the same amount of time in order
to meet eachother, so,
t%5B1%5D+=+t%5B2%5D
------------------------------
t%5B1%5D+=+d%5B1%5D+%2F+v%5B1%5D and
t%5B2%5D+=+d%5B2%5D+%2F+v%5B2%5D, therefore
d%5B1%5D+%2F+v%5B1%5D+=+d%5B2%5D+%2F+v%5B2%5D
------------------------------
d%5B1%5D+%2B+d%5B2%5D+=+11 km
If I say d%5B1%5D=+x, then d%5B2%5D+=+11+-+x
x+%2F+8+=+%2811+-+x%29+%2F+9
9x+=+88+-+8x
17x+=+88
x+=+5.1765 km
11+-+x+=+5.8235 km
The runner 1 went the shorter distance in the same time
as runner 2, but he still passed the flagpole before
runner 2, and ended up .1765 km on the west side of it.
Runner 2 met him there (6 - 5.8235) km short of the
flagpole, or .1765 km short.