document.write( "Question 757565: A started walking in north at 7 am with the speed of 5 km/hr. after one hour B started towards east on a cycle with the speed of 16 km/hr. at what time the distance between them will be 52 km? \n" ); document.write( "

Algebra.Com's Answer #460978 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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A started walking in north at 7 am with the speed of 5 km/hr.
\n" ); document.write( " after one hour B started towards east on a cycle with the speed
\n" ); document.write( " of 16 km/hr.
\n" ); document.write( " at what time the distance between them will be 52 km?
\n" ); document.write( ":
\n" ); document.write( "Let t = travel time of the bike
\n" ); document.write( "then
\n" ); document.write( "(t+1) = walking time
\n" ); document.write( ":
\n" ); document.write( "dist = speed * time
\n" ); document.write( ":
\n" ); document.write( "This is a pythag problem: a^2 + b^2 = c^2, where
\n" ); document.write( "a = 16t, bike dist
\n" ); document.write( "b = 5(t+1), walking dist
\n" ); document.write( "c = 52, dist apart in t hrs from 8 AM
\n" ); document.write( ":
\n" ); document.write( "(16t)^2 + (5(t+1)^2 = 52^2
\n" ); document.write( "256t^2 + (5t+5)^2 = 2704
\n" ); document.write( "256t^2 + 25t^2 + 50t + 25 = 2704
\n" ); document.write( "281t^2 + 50t + 25 - 2704 = 0
\n" ); document.write( "281t^2 + 50t - 2679 = 0
\n" ); document.write( "solve this equation using the quadratic formula
\n" ); document.write( "Positive solution t = 3 hrs
\n" ); document.write( "At 11 AM they will be 52 km apart
\n" ); document.write( " \n" ); document.write( "

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