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Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let T be the amount of time, in seconds, before Jenny and Kenny can see each other again. If T is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
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igloo myrtilles fourmis
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Jenny's distance is 1 foot per seccond for t seconds.
Kenny's distance is 3*J.
Assume that Jenny's vision is cyclopic-laser (that is, no parallax, no stereoscopic, and field of vision is only a half-light wave frequency in width in a flat plane.)
Jenny will only see Kenny AFTER she passes the midpoint of the circular building.
The distance after the center point is some distance s.
Jenny will walk (100+s) feet before Jenny sees Kenny; and Kenny will have walked 3*(100+s) or (300 + 3s) feet.
At the instant Jenny sees Kenny, Jenny is SquareRoot(100^2+ s^2) distance from the center point of the circular building; Kenny is SquareRoot( ( 300+3s-100))^2 + 100^2 ) from the center point; Jenny and Kenny are SquareRoot( 200^2 + ( 300+3*s - (100+s) )^2 ) apart.
With that information you have this equation:
100 / ( 200 + 3s) = s / 100
thus:
3s^2 + 200s - 100^2 = 0
s = -200 + SquareRoot( 200^2 + 4*3*100^2 ) / (2*3)
s = (-100 + 200 ) / 3
s = 100/3 '<-- distance past center, NOT TIME
John E. Franklin could you post your step-by-step solution?
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Last edited by John E. Franklin (2008-04-29 03:38:51)
igloo myrtilles fourmis
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Last edited by John E. Franklin (2008-04-29 11:48:50)
igloo myrtilles fourmis
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John E. Franklin
Show your numbers (or hide 'em).
Your calculator needs batteries!.
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