 ## Question:

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 $y_{B}$