document.write( "Question 213740: A bus driver leaves the station at 10am at 40mph. He forgets to pack luggage underneath the bus and a car is dispatched 1hr later with the forgotten luggage. If the car has only 2hrs to catch up the bus, how fast does the car need to drive? \n" ); document.write( "
Algebra.Com's Answer #161465 by josmiceli(19441)\"\" \"About 
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If the car drives 40 mi/hr or less, it can never
\n" ); document.write( "catch the bus.
\n" ); document.write( "The car leaves 1 hr after the bus has left
\n" ); document.write( "In 1 hr, the bus has gone
\n" ); document.write( "\"d+=40%2A1\"
\n" ); document.write( "\"d+=+40\" mi
\n" ); document.write( "I have a stopwatch which I will start when the car leaves
\n" ); document.write( "and I'll stop it when the car catches the bus.
\n" ); document.write( "Both vehicles will drive for 2 hrs, but the bus has
\n" ); document.write( "a \"40\" mi headstart.
\n" ); document.write( "Let \"d\" = distance the car goes in 2 hrs
\n" ); document.write( "Let \"r\" = speed of the car in mi/hr
\n" ); document.write( "\"d+-+40\" = distance the bus travels
\n" ); document.write( "Let \"t+=+2\" hrs for both
\n" ); document.write( "For car:
\n" ); document.write( "(1) \"d+=+r%2A2\"
\n" ); document.write( "For bus:
\n" ); document.write( "(2) \"d+-+40+=+40%2A2\"
\n" ); document.write( "Substitute (1) in (2)
\n" ); document.write( "\"2r+-+40+=+80\"
\n" ); document.write( "\"2r+=+120\"
\n" ); document.write( "\"r+=+60\"
\n" ); document.write( "The car needs to go 60 mi/hr to catch the bus in 2 hrs
\n" ); document.write( "check answer
\n" ); document.write( "In 2 hrs, the bus goes
\n" ); document.write( "\"d+=+40%2A2\"
\n" ); document.write( "\"d+=+80+mi\"
\n" ); document.write( "\"80+%2B+40+=+120\"
\n" ); document.write( "The car needs to go a total of 120 mi in 2 hrs
\n" ); document.write( "\"d+=+r%2At\"
\n" ); document.write( "\"120+=+r%2A2\"
\n" ); document.write( "\"r+=+60\" mi/hr
\n" ); document.write( "OK\r
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