document.write( "Question 263150: If one walks 500 feet at 4 miles per hour walking to his/her destination and walks back 500 feet at 3 miles per hour... What is the average speed for the entire trip? How do you calculate it? \n" ); document.write( "
Algebra.Com's Answer #193901 by drk(1908)\"\" \"About 
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Be careful here. Your units are not aligned - -> \"500 feet\" and \"miles\" per hour. This is an RTD question. Here is a table based on the given information:
\n" ); document.write( "destination . . . . . . . . . . . . . . rate . . . . . . . . . . . . . .time . . . . . . . . . . distance
\n" ); document.write( "to. . . . . . . . . . . . . . . . . . . . . . 4*5280 . . . . . . . . . . . . . . . . . . . . . . . . . 500
\n" ); document.write( "from. . . . . . . . . . . . . . . . . . . . 3*5280 . . . . . . . . . . . . . . . . . . . . . . . . . .500
\n" ); document.write( "average. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1000
\n" ); document.write( "we want average speed. This is average rate which is expressed as total distance over total time.
\n" ); document.write( "the time for \"to\" is 500/(4*5280) = .023674242 hrs
\n" ); document.write( "the time for \"to\" is 500/(3*5280) =. 031565656 hrs
\n" ); document.write( "so, total time = .055239898 hrs
\n" ); document.write( "and total distance is 1000 ft
\n" ); document.write( "average speed = 1000/.055239898 ft/hr ~ 18102.8571428 ft / hr
\n" ); document.write( "Now you may want the average rate in miles / hr. we get
\n" ); document.write( "18102.8571428 ft / hr * 1 mile / 5280 ft ~ 3.42857 mph. \n" ); document.write( "
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