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Jim gets on a bike and travels to a location and back. The trip is 40 miles long. The return trip is 2 hours longer and his spead is decreased by 10 miles per hour. What was Jim's speed on the trip to the location and his speed on the trip home?
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\n" ); document.write( "My understanding is there are two one-way trips of 40 mi
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\n" ); document.write( "Let the initial speed = s; Then then the returning speed = (s-10)
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\n" ); document.write( "Time = Dist/speed
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\n" ); document.write( "Time out = Time back - 2 hrs
\n" ); document.write( " 40/s = 40/(s-10) - 2
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\n" ); document.write( "Eliminate the denominators; mult equation by s(s-10); resulting in:
\n" ); document.write( " 40(s-10) = 40s - 2(s(s-10))
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\n" ); document.write( "40s - 400 = 40s - 2s^2 + 20s
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\n" ); document.write( "+2s^2 + 40s - 40s - 20s - 400 = 0
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\n" ); document.write( "2s^2 - 20s - 400 = 0; our old friend the quadratic equation
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\n" ); document.write( "s^2 - 10s - 200 = 0; simplified, divided eq by 2
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\n" ); document.write( "Factors to:
\n" ); document.write( "(s-20) (s+10) = 0
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\n" ); document.write( "Positive solution: s = +20 mph out then 10 mph back:
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\n" ); document.write( "Check using time: 40/20 = 2hr out & 40/10 = 4 hr back (2 hours longer)\r
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