document.write( "Question 267665: Lucas is 1/3 mile away from home bicycling at 20 miles per hour when his brother takes off on his bike to catch up. How fast must Lucas's brother go to catch him a mile from home. \n" ); document.write( "
Algebra.Com's Answer #196372 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Lucas is 1/3 mile away from home bicycling at 20 miles per hour when his brother takes off on his bike to catch up. \n" ); document.write( " How fast must Lucas's brother go to catch him a mile from home. \n" ); document.write( ": \n" ); document.write( "Briefly, bro has to travel 1 mi in the same time that L travels 2/3 of a mile \n" ); document.write( ": \n" ); document.write( "Let s = bro's speed to accomplish difficult feat \n" ); document.write( ": \n" ); document.write( "Write a time equation: Time = dist/speed \n" ); document.write( ": \n" ); document.write( "bro's time = L's time \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( "Cross multiply \n" ); document.write( "s = 30 mph bro has pedal to catch L 1 mi from home \n" ); document.write( "; \n" ); document.write( ": \n" ); document.write( "Check this by finding the times \n" ); document.write( "1/30 = .033 hrs \n" ); document.write( ".667/20 = .033 hrs (about 2 min) \n" ); document.write( " |