document.write( "Question 774569: THE SPEED OF A MOVING WALKWAY IS TYPICALLY ABOUT 3.6 PER SECOND. WALKING ON SUCH A MOVING WALKWAY, IT TAKES KAREN A TOTAL OF 50 SECONDS TO TRAVEL 90 FEET WITH THE MOVEMENT OF THE WALK AWAY AND THEN BACK AGAIN AGAINST THE MOVEMENT OF THE WALKAWAY. WHAT IS KAREN'S NORMAL WALKING SPEED? \n" ); document.write( "
Algebra.Com's Answer #472413 by ankor@dixie-net.com(22740)\"\" \"About 
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THE SPEED OF A MOVING WALKWAY IS TYPICALLY ABOUT 3.6 ft PER SECOND.
\n" ); document.write( " WALKING ON SUCH A MOVING WALKWAY, IT TAKES KAREN A TOTAL OF 50 SECONDS TO TRAVEL 90 FEET WITH THE MOVEMENT OF THE WALK AWAY AND THEN BACK AGAIN AGAINST THE MOVEMENT OF THE WALKAWAY.
\n" ); document.write( " WHAT IS KAREN'S NORMAL WALKING SPEED?
\n" ); document.write( ":
\n" ); document.write( "Let w = the walking speed
\n" ); document.write( "then
\n" ); document.write( "(w+3.6) = ground speed with walkway
\n" ); document.write( "and
\n" ); document.write( "(w-3.6) = ground speed against
\n" ); document.write( ":
\n" ); document.write( "Write a time equation, time = dist/speed
\n" ); document.write( ":
\n" ); document.write( "time with + time against = 50 sec
\n" ); document.write( "\"90%2F%28%28w%2B3.6%29%29\" + \"90%2F%28%28w-3.6%29%29\" = 50
\n" ); document.write( "multiply by (w-3.6)(w+3.6) results:
\n" ); document.write( "90(w-3.6) + 90(w+3.6) = 50(w-3.6)(w+3.6)
\n" ); document.write( "90w - 324 + 90w + 324 = 50(w^2 - 12.96)
\n" ); document.write( "180w = 50w^2 - 648
\n" ); document.write( "a quadratic equation
\n" ); document.write( "0 = 50w^2 - 180w - 648
\n" ); document.write( "Use the quadratic formula; a=50; b=-180; c=-648
\n" ); document.write( "I got a positive solution of 5.825 ft/sec as K's walking speed
\n" ); document.write( "See what you get
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