document.write( "Question 675711: a mover drives a moving truck. he leaves the familys house and travels at a steady rate of 40mph. the family leaves three fourths of an hour later following the same route. they travel at a steady rate of 60MPH. how long after the moving truck leaves will the car catch up?\r
\n" ); document.write( "\n" ); document.write( "i can do the equition i just dont know how to set it up. \n" ); document.write( "
Algebra.Com's Answer #419999 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "In 3/4 hr mover travels (d=rt)40*(3/4)=30 mi\r
\n" ); document.write( "\n" ); document.write( "When they both have traveled the same distance, the car will have caught up
\n" ); document.write( "Let t=time that elapses 'till they have traveled the same distance
\n" ); document.write( "distance Mover travels=30+40t
\n" ); document.write( "distance car travels =60t
\n" ); document.write( "30+40t=60t
\n" ); document.write( "20t=30
\n" ); document.write( "t=1.5 hrs----time it takes the car to catch up after the car leaves. May need to add 3/4 hr to this if we are calculating from when the truch leaves
\n" ); document.write( "CK
\n" ); document.write( "Mover travels 30+40.(1.5)=30+60=90 mi
\n" ); document.write( "Car travels 60*(1.5)=90 mi\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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