document.write( "Question 922344: How fast did a car go to overtake a passenger's bus in 5 hours if the bus' rate is 45 kph and left 4 hours before the car? \n" ); document.write( "
Algebra.Com's Answer #559551 by richwmiller(17219)\"\" \"About 
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5x=(5+4)*45
\n" ); document.write( "x=9*9
\n" ); document.write( "x=81 kph to catch it in 5 hours\r
\n" ); document.write( "\n" ); document.write( "another way\r
\n" ); document.write( "\n" ); document.write( "The first car had a head start of 180 (4*45) km
\n" ); document.write( "the second car needs to gain
\n" ); document.write( "180/5=36 km
\n" ); document.write( "so it must travel at
\n" ); document.write( "45+36=81 kph\r
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\n" ); document.write( "\n" ); document.write( "check
\n" ); document.write( "Gaining speed 81-45= 36 mph
\n" ); document.write( "Head start 45*4 =180 miles
\n" ); document.write( "Time to catch up 180/36=5 hours
\n" ); document.write( "Algebraic solution
\n" ); document.write( "45(t+4)=81t
\n" ); document.write( "45t+180=81t
\n" ); document.write( "180=81t-45t
\n" ); document.write( "180=36t
\n" ); document.write( "t=5 hours
\n" ); document.write( "Distance to meeting 5*81=405 miles
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