document.write( "Question 638443: Two contestants run a 440-yard course. The first wins by 4 seconds when given a head start of 200 feet. They finish together when the first is given a head start of 40 yards. Find the speed of each in feet per second. \r
\n" ); document.write( "\n" ); document.write( "I tried to solve by using the formula D (distance)= s (speed) x t (time).
\n" ); document.write( "But I come up with two equations for each situation. Converting yards to feet: 440yards=1320 feet:
\n" ); document.write( "runner #1: s(t-4)=1120 (wins by 4 seconds)
\n" ); document.write( "runner #2: s(t)=1320 \r
\n" ); document.write( "\n" ); document.write( "and for the second situation:
\n" ); document.write( "runner #1: s(t)=1200 (given a 40 yard=120feet head start: 1320-120=1200 feet).
\n" ); document.write( "runner #2 s(t)=1320
\n" ); document.write( "then solving for s: s=1320/t and substituting 1320/t(t-4)=1120 and solving for t, 1320t-5280=1120t, t=26.4 secs, then s=1320/t and s=1320/26.4= 50ft/sec. \r
\n" ); document.write( "\n" ); document.write( "Now, obviously a human cannot run at that speed, so this does not seem like a plausible answer, and I cannot figure out why I have four equations, I also tried to use different combinations but this didn't work either. So, I trust someone with better mathematical acumen can come up with a better solution. Thank you so kindly for trying to solve this problem. I look forward to the correct solution and appreciate your time and assistance. \r
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Algebra.Com's Answer #402200 by MathTherapy(10552)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Two contestants run a 440-yard course. The first wins by 4 seconds when given a head start of 200 feet. They finish together when the first is given a head start of 40 yards. Find the speed of each in feet per second. \r
\n" ); document.write( "\n" ); document.write( "I tried to solve by using the formula D (distance)= s (speed) x t (time).
\n" ); document.write( "But I come up with two equations for each situation. Converting yards to feet: 440yards=1320 feet:
\n" ); document.write( "runner #1: s(t-4)=1120 (wins by 4 seconds)
\n" ); document.write( "runner #2: s(t)=1320 \r
\n" ); document.write( "\n" ); document.write( "and for the second situation:
\n" ); document.write( "runner #1: s(t)=1200 (given a 40 yard=120feet head start: 1320-120=1200 feet).
\n" ); document.write( "runner #2 s(t)=1320
\n" ); document.write( "then solving for s: s=1320/t and substituting 1320/t(t-4)=1120 and solving for t, 1320t-5280=1120t, t=26.4 secs, then s=1320/t and s=1320/26.4= 50ft/sec. \r
\n" ); document.write( "\n" ); document.write( "Now, obviously a human cannot run at that speed, so this does not seem like a plausible answer, and I cannot figure out why I have four equations, I also tried to use different combinations but this didn't work either. So, I trust someone with better mathematical acumen can come up with a better solution. Thank you so kindly for trying to solve this problem. I look forward to the correct solution and appreciate your time and assistance. \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Let the speed of the 1st runner be \"S%5B1%5D\", and speed of 2nd, \"S%5B2%5D\"
\n" ); document.write( "The course is 440 yards, or 1,320 feet
\n" ); document.write( "Since the 1st gets a headstart of 200 feet, then the 1st travels 1,320 – 200, or 1,120 feet\r
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\n" ); document.write( "\n" ); document.write( "Time taken by 1st to go 1,120 feet, plus time won by, equals time taken by 2nd to go 1,320 ft, OR\r
\n" ); document.write( "\n" ); document.write( "\"1120%2FS%5B1%5D+%2B+4+=+1320%2FS%5B2%5D\"
\n" ); document.write( "
\n" ); document.write( "\"1120S%5B2%5D+%2B+4S%5B1%5DS%5B2%5D+=+1320S%5B1%5D\" ---- Multiplying by LCD, \"S%5B1%5DS%5B2%5D\" ---- eq (i)\r
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\n" ); document.write( "\n" ); document.write( "Since the 1st gets a headstart of 40 yards, or 120 feet, then the 1st travels 1,320 – 120, or 1,200 feet\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Time taken by 1st to go 1,200 feet equals time taken by 2nd to go 1,320 ft, OR\r
\n" ); document.write( "\n" ); document.write( "\"1200%2FS%5B1%5D+=+1320%2FS%5B2%5D\"
\n" ); document.write( "
\n" ); document.write( "\"1200S%5B2%5D+=+1320S%5B1%5D\" ---- Cross-multiplying ---- eq (ii)\r
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\n" ); document.write( "\n" ); document.write( "Therefore, \"1120S%5B2%5D+%2B+4S%5B1%5DS%5B2%5D+=+1200S%5B2%5D\" ---- ----- Substituting \"1200S%5B2%5D\" for \"1320S%5B1%5D\" in eq (i)
\n" ); document.write( "\"4%28280S%5B2%5D+%2B+S%5B1%5DS%5B2%5D%29+=+4%28300S%5B2%5D%29\" ----- Factoring out GCF, 4\r
\n" ); document.write( "\n" ); document.write( "\"280S%5B2%5D+%2B+S%5B1%5DS%5B2%5D+=+300S%5B2%5D\"
\n" ); document.write( "
\n" ); document.write( "\"S%5B1%5DS%5B2%5D+=+300S%5B2%5D+-+280S%5B2%5D\"
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\n" ); document.write( "\"S%5B1%5DS%5B2%5D+=+20S%5B2%5D\"\r
\n" ); document.write( "\n" ); document.write( "\"S%5B2%5D%28S%5B1%5D%29+=+S%5B2%5D%2820%29\" ---- Factoring out GCF, \"S%5B2%5D\"\r
\n" ); document.write( "\n" ); document.write( "\"S%5B1%5D++=+20\"\r
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\n" ); document.write( "\n" ); document.write( "\"S%5B1%5D\", or speed of 1st runner = \"highlight_green%2820%29\" ft/s \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"1200S%5B2%5D+=+1320%2820%29\" ---- Substituting 20 for \"S%5B1%5D\" in eq (ii)\r
\n" ); document.write( "\n" ); document.write( "\"1200S%5B2%5D+=+26400\"\r
\n" ); document.write( "\n" ); document.write( "\"S%5B2%5D\", or speed of 2nd runner = \"26400%2F1200\", or \"highlight_green%2822%29\" ft/s\r
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\n" ); document.write( "\n" ); document.write( "You can do the check!!\r
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\n" ); document.write( "\n" ); document.write( "Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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