document.write( "Question 1039115: A car's rate is 50 kph. A truck's rate is 40 kph. If the car starts later by 5 km in comparison's to the truck's, how many minutes before the car reaches the truck?\r
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document.write( "Is it 30 minutes? But how did you arrive with this answer? \n" );
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Algebra.Com's Answer #653853 by josmiceli(19441) You can put this solution on YOUR website! Start a stopwatch when the car leaves. \n" ); document.write( "Let \n" ); document.write( "hrs when the car catches up with the truck \n" ); document.write( "Let \n" ); document.write( "until it catches up with the truck \n" ); document.write( "--------------------------------------- \n" ); document.write( "Equation for the car: \n" ); document.write( "(1) \n" ); document.write( "Equation for the truck: \n" ); document.write( "(2) \n" ); document.write( "-------------------- \n" ); document.write( "Substitute (1) into (2) \n" ); document.write( "(2) \n" ); document.write( "(2) \n" ); document.write( "(2) \n" ); document.write( "------------------ \n" ); document.write( "The car catches up with the truck 1/2 after \n" ); document.write( "the CAR leaves.This is an important point. \n" ); document.write( "--------------- \n" ); document.write( "If you wanted to know the time in hrs to catch the truck \n" ); document.write( "after the truck leaves, you must add in the truck's time \n" ); document.write( "to travel 5 km \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "The car catches the truck 5/8 hr after the TRUCK leaves. \n" ); document.write( "This isn't what was wanted, so you don't care about it. \n" ); document.write( " \n" ); document.write( " |