document.write( "Question 1026488: John left work and traveled towards New York at average speed of 33 miles per hour Michael left two hours later and traveled in the same direction but with an average speed of 55 miles per hour how long did John travel until Micah caught up? \n" ); document.write( "
Algebra.Com's Answer #641817 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
John left work and traveled towards New York at average speed of 33 miles per hour
\n" ); document.write( " Michael left two hours later and traveled in the same direction but with an average speed of 55 miles per hour
\n" ); document.write( " how long did John travel until Micah caught up?
\n" ); document.write( ":
\n" ); document.write( "Let t = travel time of J when M caught up with him
\n" ); document.write( "then
\n" ); document.write( "(t-2) = travel time of M when he catches J
\n" ); document.write( ":
\n" ); document.write( "When this happens, they will both have traveled the same distance
\n" ); document.write( "Write a dist equation; d = speed * time
\n" ); document.write( "55(t-2) = 33t
\n" ); document.write( "55t - 110 = 33t
\n" ); document.write( " 55t - 33t = 110
\n" ); document.write( "22t = 110
\n" ); document.write( "t = 110/22
\n" ); document.write( "t = 5 hrs, J traveled when caught by M
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "We can confirm this solution, find the actual dist each traveled
\n" ); document.write( "Should be the same
\n" ); document.write( "55 * 3 = 165 Traveled 2 hrs less
\n" ); document.write( "33 * 5 = 165 \n" ); document.write( "
\n" );