document.write( "Question 779276: 1. Starting at noon. John drives north along a highway at a speed of 70 mi/hr until 2:30pm. How far has he travelef since noon? 2. He then continues in the same direction at a different steady speed for a distance of 75 miles until 4:00 pm. What is his speed between 2:30 pm and 4:00pm? 3. From 4:00 pm he travels an additional 180 miles in the same direction to arrive s
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Algebra.Com's Answer #475211 by AnlytcPhil(1806)\"\" \"About 
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1. Starting at noon. John drives north along a highway at a speed of 70 mi/hr until 2:30pm.
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document.write( "So at 2:30pm he has traveled 70 mi/hr for 2.5 hours. \r\n" );
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\n" ); document.write( "How far has he traveled since noon?
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document.write( "Since distance = speed×time, at 2:30pm, he has traveled 70×2.5 = 175 miles.\r\n" );
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\n" ); document.write( "2. He then continues in the same direction at a different steady speed for a distance of 75 miles until 4:00 pm.
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document.write( "So he goes 75 miles from 2:30 to 4:00pm, which is 1.5 hours.\r\n" );
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\n" ); document.write( "What is his speed between 2:30 pm and 4:00pm?
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document.write( "since speed = distance/time, his speed is 75/1.5 = 50 mi/hr  \r\n" );
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\n" ); document.write( "3. From 4:00 pm he travels an additional 180 miles in the same direction to arrive
\n" ); document.write( "at his destination. He travels this final part of his journey at a constant speed of 60 mi/hr.
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document.write( "So he travels 180 miles at 60 mi/hr,\r\n" );
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document.write( "Since time = distance/speed, 180/60 = 3 hours after 4:00pm or 7:00pm \r\n" );
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\n" ); document.write( "4. What is John's average speed between his start at noon and the time he arrives at his destination?
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document.write( "In part 1, he drives 175 miles.\r\n" );
document.write( "In part 2, he drives 75 miles.\r\n" );
document.write( "In part 3, he drives 180 miles.\r\n" );
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document.write( "So he drives a total of 175+75+180 or 430 miles\r\n" );
document.write( "From noon to 7:00 is 7 hours.\r\n" );
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document.write( "Average speed = total distance/total time = 430/7 = 61 3/7 mi/hr.\r\n" );
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document.write( "Edwin
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