document.write( "Question 55838This question is from textbook
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document.write( ": a group of people started driving at 6 mph. one of the cars didnt move until its started driving 10 mph. the car caught up to the others in 5 minutes.\r
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document.write( "question:what is the distance of the group of people while the car that was behind started to catch up? what would be the expression? \n" );
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Algebra.Com's Answer #37933 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A group of people started driving at 6 mph. One of the cars didn't move until it's started driving 10 mph. The car caught up to the others in 5 minutes. \n" ); document.write( ": \n" ); document.write( "Question: What is the distance of the group of people while the car that was behind started to catch up? What would be the expression? \n" ); document.write( ": \n" ); document.write( "Since we are dealing in minutes, change the speed to miles/min: \n" ); document.write( "6 mph = 6/60 = .1 mi/min \n" ); document.write( "10 mph = 10/60 = .167 mi/min \n" ); document.write( ": \n" ); document.write( "When the faster car catches up with the group, they will have traveled the same distance. Dist = speed * time; \n" ); document.write( " t = time the 1st group has traveled; \n" ); document.write( "(t-5) = time the late car has traveled; \n" ); document.write( ": \n" ); document.write( ".1t = .167(t-5) \n" ); document.write( ".1t = .167t - .833 \n" ); document.write( ".1t - .167t = .833 \n" ); document.write( "-.067t = -.833 \n" ); document.write( " t = -.833/-.067 \n" ); document.write( " t = 12.438 minutes when the car and the group are together \n" ); document.write( ": \n" ); document.write( "Group Distance = .1 * 12.438 = 1.24 mi at that time \n" ); document.write( " Car Distance = .167 * (12.438-5) = 1.24 mi also traveling for only 7.438 min \n" ); document.write( ": \n" ); document.write( "1.24 mi traveled when car catches up with the group \n" ); document.write( " |