SOLUTION: A river flows with a uniform velocity vr. A person in a motorboat travels 1.12 km upstream, at which time she passes a log floating by. Always with the same engine throttle setting

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Question 947523: A river flows with a uniform velocity vr. A person in a motorboat travels 1.12 km upstream, at which time she passes a log floating by. Always with the same engine throttle setting, the boater continues to travel upstream for another 1.48 km, which takes her 65.6 min. She then turns the boat around and returns downstream to her starting point, which she reaches at the same time as the same log does. How much time does the boater spend traveling back downstream?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the speed of the boat in still water
Let +c+ = the speed of the current
+s+-+c+ = the boat's speed going upstream
+s+%2B+c+ = the boat's speed going downstream
Let +t+ = time in hrs for the whole trip
-------------------------------------------
The one-way distance the boat travels is
+1.12+%2B+1.48+=+2.6+ km
the distance for the whole trip is:
+2%2A2.6+=+5.2+ km
The effect of the current cancels for the
round trip, so
+t+=+5.2%2Fs+
This is also the time for the log to travel
+1.12+ km at the speed of the current
+t+=+1.12%2Fc+
By substitution:
+5.2%2Fs+=+1.12%2Fc+
+5.2c+=+1.12s+
+c+=+.2154s+
----------------
I am also told that:
+1.48+=+%28+s+-+c+%29%2A%28+65.6%2F60+%29+
+1.48%2A%28+60%2F65.6+%29+=+s+-+c+
+s+-+c+=+1.3537+
By substitution:
+s+-+.2154s+=+1.3537+
+.7846s+=+1.3537+
+s+=+1.7253+
and, since
+c+=+.2154s+
+c+=+.2154%2A1.7253+
+c+=+.3716+
--------------
The time to go back downstream is:
+t%5B1%5D+=+2.6+%2F+%28+s+%2B+c+%29+
+t%5B1%5D+=+2.6+%2F+%28+1.7253+%2B+.3716+%29+
+t%5B1%5D+=+2.6+%2F+2.097+
+t%5B1%5D+=+1.24+ hrs
----------------------
check answer:
+2.6+=+%28+s+-+c+%29%2At%5B2%5D+
+2.6+=+%28+1.7253+-+.3716+%29%2At%5B2%5D+
+2.6+=+1.3537t%5B2%5D+
+t%5B2%5D+=+1.9207+
and
+2.6+=+%28+s+%2B+c+%29%2At%5B3%5D+
+2.6+=+%28+1.7253+%2B+.3716+%29%2At%5B3%5D+
+2.6+=+2.0969t%5B3%5D+
+t%5B3%5D+=+1.24+
+t%5B2%5D+%2B+t%5B3%5D+=+1.9207+%2B+1.24+
+t+=+3.1607+ ( this is off )
and, also,
+t+=+1.12%2Fc+
+t+=+1.12%2F.3716+
+t+=+3.014+
and also,
+t+=+5.2%2Fs+
+t+=+5.2%2F1.7253+
+t+=+3.014+
--------------------
My calculations went off somewhere
I think my method is OK
Check my math and probably get
another opinion, too