document.write( "Question 468781: One hour after a truck has left on an overnight haul, a messenger on a motorcycle leaves from the same starting point to overtake the truck. If the messenger travels at an average rate of 65 mi/h and overtakes the truck in 4 hours, what is the average rate of the truck? \n" ); document.write( "
Algebra.Com's Answer #321787 by ankor@dixie-net.com(22740)\"\" \"About 
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One hour after a truck has left on an overnight haul, a messenger on a motorcycle
\n" ); document.write( " leaves from the same starting point to overtake the truck.
\n" ); document.write( " If the messenger travels at an average rate of 65 mi/h and overtakes the truck
\n" ); document.write( " in 4 hours, what is the average rate of the truck?
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\n" ); document.write( "Let s = the speed of the truck
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\n" ); document.write( "We are given the travel time of the Motorcycle as 4 hrs
\n" ); document.write( "then
\n" ); document.write( "the travel time of the truck is one hr longer: 5 hrs
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\n" ); document.write( "When motorcycle overtakes the truck, they will have traveled the same distance
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\n" ); document.write( "Write a distance equation, dist = speed * time
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\n" ); document.write( "truck dist = motorcycle dist
\n" ); document.write( " 5s = 65(4)
\n" ); document.write( " 5s = 260
\n" ); document.write( " s = \"260%2F5\"
\n" ); document.write( " s = 52 mph is the speed of the truck
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\n" ); document.write( "see if this is true find the distance of each
\n" ); document.write( "4*65 = 260 mi
\n" ); document.write( "5*52 = 260 mi
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