 ## Question:

The man pulls the boy up to the tree limb C by walking backward at a constant speed of 1.5 m/s . Determine the speed at which the boy is being lifted at the instant ${x}_{A}$ = 4 m . Neglect the size of the limb. When ${x}_{A}$ = 0 , ${x}_{B}$ = 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 $l_{A C}$ we have $l_{A C}$=$\sqrt{x_{A}^{2}+8^{2}}$. Thus,

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

Time Derivative: realizing that $v _ { A }$ = $\frac { d x _ { A } } { d t }$ 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 $x_{A}$=4 m

$v_{B}$=-$\frac{4}{\sqrt{4^{2}+64}}$(1.5)=-0.671 m/s=0.671 m/s $\uparrow$

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