The distance (d) from the center of the seesaw varies inversely as the weight (w)
of a person. JB who weighs 50 kg sits 3 feet from the fulcrum. How far from
the fulcrum must JP sit in order to balance with JB if he weighs 35 kg?
Answers
Given : The distance (d) from the center of the seesaw varies inversely as the weight (w) of a person.
JB who weighs 50 kg sits 3 feet from the fulcrum.
JP weighs 35 kg
To Find : How far from the fulcrum must JP sit in order to balance with JB
Solution:
The distance (d) from the center of the seesaw varies inversely as the weight (w) of a person.
=> d ∝ 1/w
=> dw = k
k = constant of proportionality
JB who weighs 50 kg sits 3 feet from the fulcrum.
=> 3(50) = k
=> 150 = k
JP weighs 35 kg
=> 35 ( d) = k
=> 35 ( d) = 150
=> 7 ( d) = 30
=> d = 30/7 = 4 2/7 ft
=> d = 4.286 ft
JP must sit 30/7 ≈ 4.286 ft feet from fulcrum in order to balance with JB
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