document.write( "Question 518946: A stationary cyclist is passed by another cyclist riding with a constant speed of 3.4 m/s. Immediately as the other bike passes the stationary bicyclist hops on his bike and accelerates at 2.4m/s2 until he catches up. (a) How much time does it take to catch up and how far has he travelled in this time? \n" ); document.write( "
Algebra.Com's Answer #345366 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! A stationary cyclist is passed by another cyclist riding with a constant speed of 3.4 m/s. \n" ); document.write( " Immediately as the other bike passes the stationary bicyclist hops on his bike and accelerates at 2.4m/s2 until he catches up. \n" ); document.write( " (a) How much time does it take to catch up and how far has he traveled in this time? \n" ); document.write( ": \n" ); document.write( "Let t = time required for him to catch up \n" ); document.write( "then \n" ); document.write( "3.4t = distance (in meters) traveled by both bikes when this happens \n" ); document.write( ": \n" ); document.write( "2.4t = cyclist speed after t seconds \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( "It is equal to the other rider's dist \n" ); document.write( "1.2t^2 = 3.4t \n" ); document.write( "1.2t^2 - 3.4t = 0 \n" ); document.write( "Divide both sides by 1.2t \n" ); document.write( "1.2t(t - 2.83) = 0 \n" ); document.write( "t = 0, start time \n" ); document.write( "and \n" ); document.write( "t = 2.83 seconds to catch up \n" ); document.write( "dist = 2.83 * 3.4 = 9.622 meters \n" ); document.write( " |