document.write( "Question 1066541: From the same place at 7am,A started walking in north at the speed of 5km/hr. After 1hour B started cycling in the east at the speed of 16km/hr. At what time will they will be at distance of 52km apart from each other.
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Algebra.Com's Answer #681769 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
From the same place at 7am,A started walking in north at the speed of 5km/hr.
\n" ); document.write( " After 1hour B started cycling in the east at the speed of 16km/hr.
\n" ); document.write( " At what time will they will be at distance of 52km apart from each other.
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\n" ); document.write( "let t = time in hrs for the walker to be 52 km from the cyclist
\n" ); document.write( "and
\n" ); document.write( "(t-1) = time for the cyclist to be 52km from the walker
\n" ); document.write( "then
\n" ); document.write( "5t = the distance walked
\n" ); document.write( "and
\n" ); document.write( "16(t-1) = distance cycled
\n" ); document.write( "or we can write it as:
\n" ); document.write( "(16t-16)
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\n" ); document.write( "Using pythag
\n" ); document.write( "\"%285t%29%5E2+%2B+%2816t-16%29%5E2+=+52%5E2\"
\n" ); document.write( "\"25t%5E2+%2B+256t%5E2+-+512t+%2B+256+=+2704\"
\n" ); document.write( "combine like terms to form a quadratic equation
\n" ); document.write( "\"281t%5E2+-+512t+%2B+256+-+2704+=+0\"
\n" ); document.write( "\"281t%5E2+-+512t+-+2448+=+0\"
\n" ); document.write( "Using the quadratic formula I got a positive solution of
\n" ); document.write( "t = 4 hrs, added to 7 Am, they were 52 km away at 11 AM
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\n" ); document.write( "You can check this for yourself
\n" ); document.write( "Walker: 4*5 = 20 mi
\n" ); document.write( "Cyclist: 3*16 = 48 mi
\n" ); document.write( "dist = \"sqrt%2820%5E2%2B48%5E2%29\"\r
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