document.write( "Question 767533: Sheila leaves on a long trip driving at a steady rate of 25 miles per hour. Her sister Allison leaves from the same location traveling to the same destination 2 hours later. She drives at a steady rate of 50 miles per hour. How long after Sheila leaves home will Allison catch up?
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document.write( "I have tried this problem for approximately 3 hours. I'm in 9th grade and I'll be honest with you and say, this is a graded assignment. Thank you in advance for any help you can provide.\r
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document.write( "-Faith\r
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Algebra.Com's Answer #467669 by algebrahouse.com(1659)![]() ![]() You can put this solution on YOUR website! Distance = rate x time\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Sheila's rate = 25 mph \n" ); document.write( "Sheila's time = t \n" ); document.write( "Sheila's distance = 25t {distance = rate x time}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Allison's rate = 50mph \n" ); document.write( "Allison's time = t - 2 {she started 2 hours later or lost two hours off of the starting time} \n" ); document.write( "Allison's distance = 50(t - 2) = 50t - 100 {distance = rate x time}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "25t = 50(t - 2) {when Allison catches up with Sheila the distances are equal} \n" ); document.write( "25t = 50t - 100 {used distributive property} \n" ); document.write( "-25t = -100 {subtracted 50t from each side} \n" ); document.write( "t = 4 {divided each side by -25} \n" ); document.write( "{t corresponds with Sheila's time, which is what the question is asking for}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "In 4 hours Allison will catch up with Sheila \n" ); document.write( "{Your mistake was probably on Allison's time. You probably had t + 2.} \n" ); document.write( " For more help from me, visit: www.algebrahouse.com \n" ); document.write( " |