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The man pulls the boy up to the tree limb C by walking backward. If he starts from rest when {x}_{A} = 0 and moves backward with a constant acceleration {a}_{A} = 0.2 m/{s}^{2} , determine the speed of the boy at the instant {y}_{B} = 4 m . Neglect the size of the limb. When {x}_{A} = 0 , {y}_{B} = 8 m ,so that A and B are coincident, i.e., the rope is 16 m long .

Step-by-step

Position-Coordinate Equation: Using the Pythagorean theorem to determine l_{A C}

I=l_{A C}+y_{B}
16=\sqrt{x_{A}^{2}+8^{2}}+y_{B}
y_{B}=16-\sqrt{x_{A}^{2}+64}

Time Derivative: Taking the time derivative  v _ { A } = \frac { d x _ { A } } { d t } and v_{B}=\frac{d y_{B}}{d t}, we have

v_{B}=\frac{d y_{B}}{d t}=-\frac{x_{A}}{\sqrt{x_{A}^{2}+64}} \frac{d x_{A}}{d t}
v_{B}=-\frac{x_{A}}{\sqrt{x_{A}^{2}+64}} v_{A}

At the instant y_{B}=4 m, 4=16-\sqrt{x_{A}^{2}+64}, x_{A}=8.944 m . The
velocity of the man at that instant can be obtained.

v_{A}^{2}=\left(v_{0}\right)_{A}^{2}+2\left(a_{c}\right)_{A}\left[s_{A}-\left(s_{0}\right)_{A}\right]
v_{A}^{2}=0+2(0.2)(8.944-0)
v_{A}=1.891 m/s

v_{B}=-\frac{8.944}{\sqrt{8.944^{2}+64}}(1.891)=-1.41 m/s=1.41 m/s\uparrow

Note: The negative sign indicates that velocity v_{B} is in the opposite direction to that of positive y_{B}

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