SOLUTION: Trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 40 MPH and Train B is Traveling at 50 MPH. Train A passes a station at 3:10 P.M. If T

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Question 183239: Trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 40 MPH and Train B is Traveling at 50 MPH. Train A passes a station at 3:10 P.M. If Train B passes the same station at 3:22 P.M. when will Train B catch Train A?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Each train has it's own equation
d%5BA%5D+=+r%5BA%5D%2At%5BA%5D
and
d%5BB%5D+=+r%5BB%5D%2At%5BB%5D
given:
r%5BA%5D+=+40 mi/hr
r%5BB%5D+=+50 mi/hr
Assume I have a stopwatch, and I start it when
when train B passes the station 12 min, or 12%2F60
of an hour after train A has passed. I will stop
the stopwatch when they meet.
How much of a headstart does train A have when B
passes the station? Let d%5Bh%5D = headstart distance
d%5Bh%5D+=+r%5BA%5D%2At%5Bh%5D
d%5Bh%5D+=+40%2A%2812%2F60%29
d%5Bh%5D+=+8mi
the distance A has to travel compared to B is:
d%5BA%5D+=+d%5BB%5D+-+8mi
Their time of travel will be the same
t%5BA%5D+=+t%5BB%5D (I'll call it t)
------------------
I can rewrite the equations
d%5BA%5D+=+r%5BA%5D%2At%5BA%5D
d%5BB%5D+=+r%5BB%5D%2At%5BB%5D
----------------------
(1) d%5BB%5D+-+8+=+40t
(2) d%5BB%5D+=+50t
This is 2 equations and 2 unknowns so it's solvable
(1) d%5BB%5D+-+40t+=+8
(2) d%5BB%5D+-+50t+=+0
Subtract (2) from (1)
(1) d%5BB%5D+-+40t+=+8
(2) -d%5BB%5D+%2B+50t+=+0
(3) 10t+=+8
(3) t+=+4%2F5 hr (or .8%2A60+=+48min)
This is the time that passes since train B goes through
station at 3:22 PM 22+%2B+48+=+70
The trains will meet at 4:10 PM
check answer:
(1) d%5BB%5D+-+8+=+40t
(2) d%5BB%5D+=+50t
------------------------
(1) d%5BB%5D+-+8+=+40%2A%284%2F5%29
(1) d%5BB%5D+=+32+%2B+8
(1) d%5BB%5D+=+40 mi
(2) d%5BB%5D+=+50%2A%284%2F5%29
(2) d%5BB%5D+=+40 mi
OK