SOLUTION: A train leaves San Diego at 3:00 PM. A second train leaves the same city in the same direction at 7:00 PM. The second train travels 56 mph faster than the first. If the second trai

Algebra ->  Average -> SOLUTION: A train leaves San Diego at 3:00 PM. A second train leaves the same city in the same direction at 7:00 PM. The second train travels 56 mph faster than the first. If the second trai      Log On


   

Question 1017189: A train leaves San Diego at 3:00 PM. A second train leaves the same city in the same direction at 7:00 PM. The second train travels 56 mph faster than the first. If the second train overtakes the first at 10:00 PM, what is the speed of each of the two trains?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the speed in mi/hr of the 1st train
+s+%2B+56+ = the speed in mi/hr of the 2nd train
---------------------
What is the head start of the 1st train in miles?
+d%5B1%5D+=+s%2A4+ ( 3PM to 7 PM is 4 hrs )
---------------------
Let +d+ = distance in miles the 2nd train
travels until it catches up with 1st train
+t+ = time in hrs from 7 PM to 10 PM, so
+t+=+3+ hrs
-------------------------------------
Equation for 1st train:
(1) +d+-+4s+=+s%2A3+
Equation for 2nd train:
(2) +d+=+%28+s+%2B+56+%29%2A3+
--------------------
(1) +d+=+3s+%2B+4s+
(1) +d+=+7s+
and
(2) +d+=+3s+%2B+168+
--------------------
+7s+=+3s+%2B+168+
+4s+=+168+
+s+=+42+
and
+s+%2B+56+=+98+
---------------
The 1st train's speed is 42 mi/hr
The 2nd train's speed is 98 mi/hr
----------------------------
check:
(1) +d+-+4s+=+s%2A3+
(1) +d+-+4%2A42+=+42%2A3+
(1) +d+-+168+=+126+
(1) +d+=+294+ mi
and
(2) +d+=+%28+s+%2B+56+%29%2A3+
(2) +d+=+%28+42+%2B+56+%29%2A3+
(2) +d+=+294+ mi
OK