document.write( "Question 900265: a man and his dog running on a straight beach.at a given point in time the dog is 12m from his owner who starting in a direction perpendicular to the beach with a certain constant speed. the dog runs twice as fast and always to wards his owner where do the man and his dog meet? \n" ); document.write( "
Algebra.Com's Answer #546046 by ankor@dixie-net.com(22740)\"\" \"About 
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a man and his dog running on a straight beach.at a given point in time the dog is 12m from his owner who starting in a direction perpendicular to the beach with a certain constant speed.
\n" ); document.write( " the dog runs twice as fast and always to wards his owner where do the man and his dog meet?
\n" ); document.write( ":
\n" ); document.write( "speed and distance vary directly
\n" ); document.write( "let d = distance the man travels from the beach,when the dog reaches him
\n" ); document.write( "then
\n" ); document.write( "2d = distance run by the dog
\n" ); document.write( ":
\n" ); document.write( "Actually the dog would probably run a curved path looking at the owner as he ran.
\n" ); document.write( "We will assume the dog ran to a point where the man is, a pythag problem
\n" ); document.write( "a^2 + b^2 = c^2, where
\n" ); document.write( "a = d
\n" ); document.write( "b = 12
\n" ); document.write( "c = 2d
\n" ); document.write( "d^2 + 12^2 = (2d)^2
\n" ); document.write( "d^2 + 144 = 4d^2
\n" ); document.write( "144 = 4d^2 - d^2
\n" ); document.write( "144 = 3d^2
\n" ); document.write( "d^2 = 144/3
\n" ); document.write( "d^2 = 48
\n" ); document.write( "d = \"sqrt%2848%29\"
\n" ); document.write( "d = \"4sqrt%283%29\" or about 6.93 meters
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