document.write( "Question 395255: Can you help? Train A&B traveling in the same direction, train A is traveling at 60 miles per hr and train B is traveling at 70 miles per hr, train A passes a station at 3:25 p.m. and train B passes the same station at 3:55 p.m. at what time will train B catch up with train A? Thank you! Shelly \n" ); document.write( "
Algebra.Com's Answer #280599 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Train A&B traveling in the same direction, train A is traveling at 60 miles per \n" ); document.write( " hr and train B is traveling at 70 miles per hr, train A passes a station at 3:25 \n" ); document.write( " p.m. and train B passes the same station at 3:55 p.m. \n" ); document.write( " at what time will train B catch up with train A? \n" ); document.write( ": \n" ); document.write( "From the given information, we know at 3:25 pm, train B is 30 min behind train A \n" ); document.write( ": \n" ); document.write( "Since it will take .5 hr to reach the station B must be 70*.5 = 35 mi behind A \n" ); document.write( ": \n" ); document.write( "let t = travel time for B to catch A \n" ); document.write( ": \n" ); document.write( "Write a distance equation: dist = speed * time \n" ); document.write( ": \n" ); document.write( "B's travel dist = A's travel dist + 35 mi \n" ); document.write( "70t = 60t + 35 \n" ); document.write( "70t - 60t = 35 \n" ); document.write( "10t = 35 \n" ); document.write( "t = \n" ); document.write( "t = 3.5 hrs for B to catch A \n" ); document.write( "then \n" ); document.write( "3:25 + 3.5 = 6:55 PM B catches A \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this by finding the distance each traveled \n" ); document.write( "70 * 3.5 = 245 mi \n" ); document.write( "60 * 3.5 = 210 mi \n" ); document.write( "----------------- \n" ); document.write( "difference: 35 mi \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Did this make sense to you? \n" ); document.write( " |