document.write( "Question 551725: Sara leaves at 7am at a rate of 45mph
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document.write( "Mark leaves at 7:30am at a rate of 55mph
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document.write( "They both leave from the same location traveling in the same direction
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document.write( "At what time will Mark catch up with Sara?\r
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document.write( "What equation or equations should be used when solving this problem algebraically? \n" );
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Algebra.Com's Answer #359858 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Sara leaves at 7am at a rate of 45mph \n" ); document.write( "Mark leaves at 7:30am at a rate of 55mph \n" ); document.write( "They both leave from the same location traveling in the same direction \n" ); document.write( "At what time will Mark catch up with Sara? \n" ); document.write( "What equation or equations should be used when solving this problem algebraically? \n" ); document.write( ": \n" ); document.write( "Let t = Sara's travel time \n" ); document.write( "then \n" ); document.write( "(t-.5) = Mark's travel time (leaves half hour later) \n" ); document.write( ": \n" ); document.write( "A simple equation can be derived from the fact, when M catches S, they will have traveled the same distance, \n" ); document.write( "Dist = speed * time \n" ); document.write( ": \n" ); document.write( "M's dist = S's dist \n" ); document.write( "55(t-.5) = 45t \n" ); document.write( "55t - 27.5 = 45t \n" ); document.write( "55t - 45t = 27.5 \n" ); document.write( "10t = 27.5 \n" ); document.write( "t = \n" ); document.write( "t = 2.75 hrs from S's starting time (2 hrs + .75(60) = 45 min) \n" ); document.write( ": \n" ); document.write( "7:00 + 2:45 = 9:45 when M catches S \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "We can confirm this by finding the dist each traveled, if we did this right, they should be equal \n" ); document.write( "45*2.75 = 123.75 mi \n" ); document.write( "55*2.25 = 123.75 \n" ); document.write( " |