document.write( "Question 178157: a woman driving a car 12ft long is passing a truck 32 ft long. the truck is traveling at 50mi/h. how fast must the woman drive her car so that she can pass the truck completely in 6 sec. (Use feet and seconds instead of miles and hours.) \n" ); document.write( "
Algebra.Com's Answer #133103 by Alan3354(69443)\"\" \"About 
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a woman driving a car 12ft long is passing a truck 32 ft long. the truck is traveling at 50mi/h. how fast must the woman drive her car so that she can pass the truck completely in 6 sec. (Use feet and seconds instead of miles and hours.)
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\n" ); document.write( "The distance from the point where the car's front is adjacent to the truck's rear to where the car's rear is adjacent to the front of the truck is:
\n" ); document.write( "12+32+12 feet = 56 feet.
\n" ); document.write( "To travel that distance in 6 seconds she must go 50 mph + 56/6 ft/sec.
\n" ); document.write( "To convert mph to ft/sec, multiply it by 22/15 (you can work that out or take my word for it).
\n" ); document.write( "So her speed must be 50*(22/15) + 56/6 feet/second
\n" ); document.write( "= 1100/15 + 28/3
\n" ); document.write( "= 220/3 + 28/3
\n" ); document.write( "= 248/3 feet/second, or 82.667 fps
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\n" ); document.write( "For mph, that's (248/3)*(15/22)
\n" ); document.write( "= 1240/22
\n" ); document.write( "= 56.36 mph. \n" ); document.write( "
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