SOLUTION: an engine pulls a train 140 miles. then a second engine, whose average rate is 5 miles per hour faster than the first engine, takes over and pulls the train 200 miles. The total ti

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Question 231501: an engine pulls a train 140 miles. then a second engine, whose average rate is 5 miles per hour faster than the first engine, takes over and pulls the train 200 miles. The total time required for both engines is 9 hours. Find the average rate of each engine.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Write a separate equation for the 2 engines
1st engine:
(1) d%5B1%5D+=+r%5B1%5D%2At%5B1%5D
2nd engine:
(2) d%5B2%5D+=+r%5B2%5D%2At%5B2%5D
----------------------
given:
d%5B1%5D+=+140 mi
d%5B2%5D+=+200 mi
r%5B2%5D+=+r%5B1%5D+%2B+5 mi/hr
t%5B1%5D+%2B+t%5B2%5D+=+9 hrs
----------------------
Now I can rewrite the equations
(1) d%5B1%5D+=+r%5B1%5D%2At%5B1%5D
(1) 140+=+r%5B1%5D%2At%5B1%5D
and
(2) d%5B2%5D+=+r%5B2%5D%2At%5B2%5D
(2) 200+=+r%5B2%5D%2A%289+-+t%5B1%5D%29
(2) 200+=+%28r%5B1%5D+%2B+5%29%2A%289+-+t%5B1%5D%29
-----------------------
At this point, I have 2 equations and
2 unknowns, so it's solvable
(2) 200+=+%28r%5B1%5D+%2B+5%29%2A%289+-+t%5B1%5D%29
(2) 200+=+9r%5B1%5D+%2B+45+-+r%5B1%5Dt%5B1%5D+-+5t%5B1%5D
Now I substitute (1) in (2)
(2) 200+=+9r%5B1%5D+%2B+45+-+140+-+5t%5B1%5D
(2) 5t%5B1%5D+=+9r%5B1%5D+-+295
Now I can substitute r[1] from (1) in (2)
(2) 5t%5B1%5D+=+9%2A%28140%2Ft%5B1%5D%29+-+295
Multiply both sides by t%5B1%5D
(2) 5%28t%5B1%5D%29%5E2+=+1260+-+295t%5B1%5D
(2) 5%28t%5B1%5D%29%5E2+%2B+295t%5B1%5D+-+1260+=+0
(2) t%5B1%5D%5E2+%2B+59t%5B1%5D+-+252+=+0
Solve using quadratic formula
t%5B1%5D+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a+=+1
b+=+59
c+=+-252
t%5B1%5D+=+%28-59+%2B-+sqrt%28+59%5E2-4%2A1%2A%28-252%29+%29%29%2F%282%2A1%29+
t%5B1%5D+=+%28-59+%2B-+sqrt%28+3481+%2B+1008%29+%29%29%2F2+
t%5B1%5D+=+%28-59+%2B-+sqrt%28+4489%29+%29%29%2F2+
t%5B1%5D+=+%28-59+%2B-+67+%29%29%2F2+ (reject the negative solution)
t%5B1%5D+=+8%2F2
t%5B1%5D+=+4 hrs
and, since
t%5B1%5D+%2B+t%5B2%5D+=+9
t%5B2%5D+=+9+-+4
t%5B2%5D+=+5 hrs
From (1)
(1) 140+=+r%5B1%5D%2At%5B1%5D
140+=+r%5B1%5D%2A4
r%5B1%5D+=+140%2F4
r%5B1%5D+=+35 mi/hr
From (2)
200+=+r%5B2%5D%2At%5B2%5D
200+=+r%5B2%5D%2A5
r%5B2%5D+=+200%2F5
r%5B2%5D+=+40 mi/hr
The average rate of the engines is 35 mi/hr and 40 mi/hr
check:
(2) 200+=+%28r%5B1%5D+%2B+5%29%2A%289+-+t%5B1%5D%29
200+=+%2835+%2B+5%29%2A%289+-+4%29
200+=+40%2A5
200+=+200
OK