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The girl at C stands near the edge of the pier and pulls in the rope horizontally at a constant speed of 6 ft/s . Determine how fast the boat approaches the pier at the instant the rope length AB is 50 ft.

Step-by-step

The length l of cord is
\sqrt{(8)^{2}+x_{B}^{2}}+x_{C}=l
Taking the time derivative:
\frac{1}{2}\left[(8)^{2}+x_{B}^{2}\right]^{-1 / 2} 2x_{B}\dot{x}_{B}+\dot{x}_{C}=0
\dot{x}_{C}=6 ft/s
When A B=50 ft
x_{B}=\sqrt{(50)^{2}-(8)^{2}}=49.356 ft

\frac{1}{2}\left[(8)^{2}+(49.356)^{2}\right]^{-1 / 2} 2(49.356)\left(\dot{x}_{B}\right)+6=0
\dot{x}_{B}=-6.0783=6.08 ft/s \leftarrow

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