document.write( "Question 266995: When Caitlin starts running towards Asha, Asha is 100 meters from Caitlin. As Caitlin runs, Asha moves directly away from Caitlin at one-third of Caitlin’s speed. How far has Caitlin run when she first catches up to Asha?\r
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Algebra.Com's Answer #196150 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! When Caitlin starts running towards Asha, Asha is 100 meters from Caitlin. \n" ); document.write( " As Caitlin runs, Asha moves directly away from Caitlin at one-third of Caitlin’s speed. \n" ); document.write( "How far has Caitlin run when she first catches up to Asha? \n" ); document.write( ": \n" ); document.write( "Briefly, C has to run 100 m more than A in the same amt of time \n" ); document.write( ": \n" ); document.write( "Let s = A's running speed \n" ); document.write( "then \n" ); document.write( "3s = C's running speed \n" ); document.write( ": \n" ); document.write( "Let d = distance run by A \n" ); document.write( "then \n" ); document.write( "(d+100) = distance run by C \n" ); document.write( ": \n" ); document.write( "Write a time equation: Time = dist/speed \n" ); document.write( ": \n" ); document.write( "C's running time = A's running time \n" ); document.write( " \n" ); document.write( "Cross multiply \n" ); document.write( "s(d+100) = 3s(d) \n" ); document.write( "sd + 100s = 3sd \n" ); document.write( ": \n" ); document.write( "Divide thru by s and you have: \n" ); document.write( "d + 100 = 3d \n" ); document.write( "100 = 3d - d \n" ); document.write( "100 = 2d \n" ); document.write( "d = 50m is the distance run by A \n" ); document.write( "then \n" ); document.write( "50 + 100 = 150m is the distance run by C to catch A \n" ); document.write( ": \n" ); document.write( "Check this, assume A's speed is 10 m/sec, C's speed is 30 m/sec \n" ); document.write( "Check the times of each, they should be equal \n" ); document.write( "150/30 = 5 min \n" ); document.write( "50/10 = 5 min \n" ); document.write( " |