document.write( "Question 1208621: A boat travels 30 km/h in still water. If this boat goes upstream in a river flowing at a rate of 12 km/h, for 5 km, and returns back downstream to its starting point, what is the average speed for the whole trip in km/h? \n" ); document.write( "
Algebra.Com's Answer #846992 by ikleyn(52864)\"\" \"About 
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\n" ); document.write( "A boat travels 30 km/h in still water. If this boat goes upstream in a river flowing at a rate of 12 km/h, for 5 km,
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document.write( "The boat's own speed in still water  is 30 km/h.\r\n" );
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document.write( "The rate of the current in the river is 12 km/h.\r\n" );
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document.write( "So,  the boat' speed with the current (downstream)  is  u = 30 + 12 = 42 km/h;\r\n" );
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document.write( "     the boat' speed against the current (upstream) is   v = 30 - 12 = 18 km/h.\r\n" );
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document.write( "Now, if you know the theory, you may use the ready formula for the average speed of the two-ways trip\r\n" );
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document.write( "    \"V%5Baverage%5D\" = \"%282u%2Av%29%2F%28u%2Bv%29\" = \"%282%2A42%2A18%29%2F%2842%2B18%29\" = \"%282%2A42%2A18%29%2F60\" = \"2%2A42%2A%283%2F10%29\" = 25.2 km/h.    ANSWER\r\n" );
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document.write( "        If you don't know this theory, \r\n" );
document.write( "        you may complete the solution in couple of steps.\r\n" );
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document.write( "The travel time upstream is   \"5%2F18\"  of an hour;  travel time downstream is  \"5%2F42\"  of an hour.\r\n" );
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document.write( "The average speed is  the total distance 5 + 5 = 10 km, divided by the total time  \"5%2F18\" + \"5%2F42\"  of an hour.\r\n" );
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document.write( "Thus the average speed is\r\n" );
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document.write( "    \"10%2F%285%2F18%2B5%2F42%29\" = \"10%2F%28%28%285%2A42%2B5%2A18%29%2F%2818%2A42%29%29%29\" = \"%2810%2A%2818%2A42%29%29%2F%285%2A42%2B5%2A18%29\" = \"%282%2A18%2A42%29%2F%2842%2B18%29\" = the same formula and the seme result = 25.2 km/h.\r\n" );
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document.write( "ANSWER.  The average speed of the two-ways trip is 25.2 km.h\r\n" );
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