document.write( "Question 803349: A truck and a car driving uniformly from city A to city B. The truck leaves at 9:15A.M and arrives at 1:15P.M. The car leaves at 10:00A.M and arrives at 12:45P.M. At what time does the car overtake the truck? \n" ); document.write( "
Algebra.Com's Answer #484422 by ankor@dixie-net.com(22740)\"\" \"About 
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A truck and a car driving uniformly from city A to city B.
\n" ); document.write( " The truck leaves at 9:15A.M and arrives at 1:15P.M.
\n" ); document.write( " The car leaves at 10:00A.M and arrives at 12:45P.M.
\n" ); document.write( " At what time does the car overtake the truck?
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\n" ); document.write( "Truck took 4 hrs to make the trip, Car took 2.75 hrs
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\n" ); document.write( "Let d = dist from A to B
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\n" ); document.write( "Find the speed of the two vehicles (speed = dist/time)
\n" ); document.write( "\"d%2F4\" = speed of truck
\n" ); document.write( "\"d%2F2.75\" = speed of the car
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\n" ); document.write( "let t = time for the car to overtake the truck
\n" ); document.write( "The truck leaves 45 min before the car, therefore
\n" ); document.write( "(t+.75) = travel time of the truck
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\n" ); document.write( "When the car overtakes the truck they will have traveled the same distance.
\n" ); document.write( "Write a distance equation dist = speed *time
\n" ); document.write( "Car dist = truck dist
\n" ); document.write( "\"d%2F2.75\"t = \"d%2F4\"(t+.75)
\n" ); document.write( "CRoss multiply
\n" ); document.write( "4dt = 2.75d(t+.75)
\n" ); document.write( "4dt = 2.75dt + 2.0625d
\n" ); document.write( "get rid of d, divide eq by d
\n" ); document.write( "4t = 2.75t + 2.0625
\n" ); document.write( "4t - 2.75t = 2.0625
\n" ); document.write( "1.25t = 2.0625
\n" ); document.write( "t = 2.0625/1.25
\n" ); document.write( "t = 1.65 hrs for the car to overtake the truck
\n" ); document.write( "Which is 1 + .65(60) = 1 hr 39 min
\n" ); document.write( "The car overtakes the truck at: 10:00 + 1:39 = 11:39 AM
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\n" ); document.write( "See if they travel the same distance in that time
\n" ); document.write( "Assume the A-B distance is 200 mi
\n" ); document.write( "then
\n" ); document.write( "\"200%2F4\" = 50 mph is the truck speed
\n" ); document.write( "and
\n" ); document.write( "\"200%2F2.75\" = 72.7 mph is the car speed
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\n" ); document.write( "Find the distances
\n" ); document.write( "Truck 50(1.65+.75) = 120 mi
\n" ); document.write( "Car: 72.7(1.65) ~ 120 mi; confirms our solution\r
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