document.write( "Question 753843: Two cars A and B starts from rest at the same time and same point. Car A moves north at 5 m/second square for 5 sec and then moves at uniform velocity. Car B moves south accelerating 4m/second square for 8 second and then moves at uniform velocity. Calculate the distance between both the car after 20seconds \n" ); document.write( "
| Algebra.Com's Answer #458668 by Alan3354(69443)     You can put this solution on YOUR website! Two cars A and B starts from rest at the same time and same point. Car A moves north at 5 m/second square for 5 sec and then moves at uniform velocity. Car B moves south accelerating 4m/second square for 8 second and then moves at uniform velocity. Calculate the distance between both the car after 20seconds \n" ); document.write( "---------------- \n" ); document.write( "Car A's distance for the 1st 5 seconds = at^2/2 = 5*5/2 = 12.5 meters \n" ); document.write( "Then it's going a*t = 5*5 m/sec for 15 seconds = 375 meters. \n" ); document.write( "In 20 seconds it goes 387.5 meters. \n" ); document.write( "-------------- \n" ); document.write( "Car B: \n" ); document.write( "at^2/2 = 4*8/2 = 16 meters \n" ); document.write( "v = at = 4*8 = 32 m/sec for the remaining 12 seconds \n" ); document.write( "s = 16 + 32*12 = 400 meters \n" ); document.write( "------- \n" ); document.write( "Add the distances when going in opposite directions \n" ); document.write( "s = 787.5 meters \n" ); document.write( "===================== \n" ); document.write( "PS Acceleration is m/sec/sec, not m/sec. \n" ); document.write( " \n" ); document.write( " |