document.write( "Question 213169This question is from textbook Prentice Hall Algebra I Louisiana
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document.write( ": A moving van leaves a house traveling at an average rate of 35mph. The family leaves the house in a car 3/4 hour later traveling the same route at an average rate of 50mph. How long will it take the family to catch up with the van? \n" );
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Algebra.Com's Answer #161056 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A moving van leaves a house traveling at an average rate of 35mph. \n" ); document.write( " The family leaves the house in a car 3/4 hour later traveling the same route at an average rate of 50mph. \n" ); document.write( "How long will it take the family to catch up with the van? \n" ); document.write( ": \n" ); document.write( "Change 3/4 hr to .75 hr \n" ); document.write( ": \n" ); document.write( "Let t = time required by the family to catch up \n" ); document.write( "then \n" ); document.write( "(t+.75) = Van travel time when it is caught \n" ); document.write( ": \n" ); document.write( "When this happens, they will have traveled the same distance \n" ); document.write( "Write a distance equation: dis = speed * time \n" ); document.write( ": \n" ); document.write( "Car dist = Van dist \n" ); document.write( "50t = 35(t+.75) \n" ); document.write( "50t = 35t + 26.25 \n" ); document.write( "50t - 35t = 26.25 \n" ); document.write( "15t = 26.25 \n" ); document.write( "t = \n" ); document.write( "t = 1.75 hrs for the family to catch the van \n" ); document.write( "; \n" ); document.write( ": \n" ); document.write( "Check solution by ensuring that they traveled the same dist: \n" ); document.write( "50 * 1.75 = 87.5 mi \n" ); document.write( "35 * (1.75+.75) = 87.5 \n" ); document.write( " |