document.write( "Question 495931: Car \"A\" is travelling at a speed of 40MPH.
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document.write( "Car \"B\" is 170 yards in front.
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document.write( "Both cars travel for 530 yards and during that time car \"A\" reaches speeds of 70MPH to catch upto car \"B\".
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document.write( "What speed is car \"B\" travelling at? \n" );
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Algebra.Com's Answer #336307 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! Car \"A\" is traveling at a speed of 40MPH. \n" ); document.write( "Car \"B\" is 170 yards in front. \n" ); document.write( "Both cars travel for 530 yards and during that time car \"A\" reaches speeds of 70MPH to catch up to car \"B\". \n" ); document.write( "What speed is car \"B\" traveling at? \n" ); document.write( ": \n" ); document.write( "What bothers me about this problem is that it says both cars travel 530 yds \n" ); document.write( "Obviously for Car A to catch up with car B, it will have travel further than B \n" ); document.write( ": \n" ); document.write( "When A catches B \n" ); document.write( "Car B travels 530 yds \n" ); document.write( "Car A travels 530 + 170 = 700 yds \n" ); document.write( ": \n" ); document.write( "Find the average velocity of Car A that uniformly accelerates from 40 to 70 mph \n" ); document.write( "Av = \n" ); document.write( "Av = 55 mph \n" ); document.write( ": \n" ); document.write( "Let s = speed of car B \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( "700s = 55 * 530 \n" ); document.write( "700s = 29150 \n" ); document.write( "s = \n" ); document.write( "s = 41.643 mph is Car B's speed \n" ); document.write( " |