SOLUTION: A train travelling from Ay town to Bee town meets with an accident after 1 hour. The train is stopped for 30 minutes, after which it proceeds at four-fifths of its usual rate, arri

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Question 696510: A train travelling from Ay town to Bee town meets with an accident after 1 hour. The train is stopped for 30 minutes, after which it proceeds at four-fifths of its usual rate, arriving at Bee town 2 hours late. If the train had covered 80 miles more before the accident, it would have just been one hour late. What is the usual rate of the train?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Equation for normal trip from A to B is
(1) +d+=+r%2At+
In +1+ hr, the train covers +d%5B1%5D+=+r%2A1+
The train is stopped for 30 minutes
+1.5+ hrs have now passed
Now it's rate is +%284%2F5%29%2Ar+ for the remaining time of
+t+%2B+2+-+1.5++=+t+%2B+.5+
The distance covered is +d%5B2%5D+=+%284%2F5%29%2Ar%2A%28+t+%2B+.5+%29+
Now I have +d+=+d%5B1%5D+%2B+d%5B2%5D+
(2) +d+=+r+%2B+%284%2F5%29%2Ar%2A%28+t+%2B+.5+%29+
So far I have 2 equations with 3 unknowns, so
I need 1 more equation
---------------------
If the train covered +80+ mi at rate = +r+,
it traveled +80+=+r%2At%5B1%5D+
+t%5B1%5D+=+80%2Fr+
Now the trip takes +t+%2B+1+ hrs, so the rest of the trip takes
+t+%2B+1+-+80%2Fr+
The train is stopped for 30 minutes, so
There is +t+%2B+1+-+.5+-+80%2Fr+ hrs left
The distance covered after the accident is then
+d%5B3%5D+=+%284%2F5%29%2Ar%2A%28+t+%2B+.5+-+80%2Fr+%29+, and
(3) +d+=+80+%2B+%284%2F5%29%2Ar%2A%28+t+%2B+.5+-+80%2Fr+%29+
-----------------------------------
Substitute (1) into (2)
(1) +d+=+r%2At+
(2) +d+=+r+%2B+%284%2F5%29%2Ar%2A%28+t+%2B+.5+%29+
(2) +r%2At+=+r+%2B+%284%2F5%29%2Ar%2A%28+t+%2B+.5+%29+
(2) +5r%2At+=+5r+%2B+4%2Ar%2A%28+t+%2B+.5+%29+
(2) +5r%2At+=+5r+%2B+4r%2At+%2B+2r+
(2) +r%2At+=+7r+
(2) +t+=+7+
-------------
(3) +d+=+80+%2B+%284%2F5%29%2Ar%2A%28+t+%2B+.5+-+80%2Fr+%29+
(3) +d+=+80+%2B+%284%2F5%29%2Ar%2A%28+7+%2B+.5+-+80%2Fr+%29+
(3) +d+=+80+%2B+%284%2F5%29%2A%28+7.5r+-+80+%29+
(3) +d+=+80+%2B+6r+-+64+
(3) +d+=+16+%2B+6r+
and since
+d+=+r%2At+
+d+=+7r+
(3) +7r+=+16+%2B+6r+
(3) +r+=+16+
The usual rate of the train is 16 mi/hr
check answer:
(1) +d+=+r%2At+
(1) +d+=+16%2A7+
(1) +d+=+112+ mi
-----------------
(2) +d+=+r+%2B+%284%2F5%29%2Ar%2A%28+t+%2B+.5+%29+
(2) +d+=+16+%2B+%284%2F5%29%2A16%2A%28+7+%2B+.5+%29+
(2) +d+=+16+%2B+%284%2F5%29%2A16%2A7.5+
(2) +d+=+16+%2B+.8%2A120+
(2) +d+=+16+%2B+96+
(2) +d+=+112+ mi
-----------------
(3) +d+=+80+%2B+%284%2F5%29%2Ar%2A%28+t+%2B+.5+-+80%2Fr+%29+
(3) +d+=+80+%2B+.8%2A16%2A%28+7+%2B+.5+-+80%2F16+%29+
(3) +d+=+80+%2B+.8%2A16%2A%28+7+%2B+.5+-+5+%29+
(3) +d+=+80+%2B+.8%2A16%2A2.5+
(3) +d+=+80+%2B+32+
(3) +d+=+112+ mi
OK