## Question:

The roller at A is moving upward with a velocity of ${v}_{A}$ = 3 ft/s and has an acceleration of ${a}_{A}$ = 4 ft/${s}^{2}$ when ${s}_{A}$ = 4 ft . Determine the velocity and acceleration of block B at this instant.

## Step-by-step

$s_{B}$+$\sqrt{\left(s_{A}\right)^{2}+3^{2}}$=l
$\dot{s}_{B}$+$\frac{1}{2}\left[\left(s_{A}\right)^{2}+3^{2}\right]^{-\frac{1}{2}}$ $\left(2 s_{A}\right)$$\dot{s}_{A}$=0 $\dot{s_{B}}$+$\left[s_{A}^{2}+9\right]^{-\frac{1}{2}}\left(s_{A} \dot{s}_{A}\right)$=0
$\ddot{s}_{B}$$\left[\left(s_{A}\right)^{2}+9\right]^{-\frac{3}{1}}\left(s_{A}^{2} \dot{s}_{A}^{2}\right)$+$\left[s_{A}^{2}+9\right]^{-\frac{1}{2}}\left(\dot{s}_{A}^{2}\right)$+$\left[s_{A}^{2}+9\right]^{-1}\left(s_{A} \ddot{s}_{A}\right)$=0
At $s_{A}$=4 ft, $\dot{s}_{A}$=3 ft/s $\ddot{s}_{A}$=4 ft/${s}^{2}$
$\dot{s}_{B}$+$\left(\frac{1}{5}\right)$(4)(3)=0
$v_{B}$=-2.4 ft/s=2.40 ft/s $\rightarrow$
${s}_{B}$$\left(\frac{1}{5}\right)^{3}(4)^{2}(3)^{2}$+$\left(\frac{1}{5}\right)(3)^{2}$+$\left(\frac{1}{5}\right)$(4)(4)=0
$a_{B}$=-3.85 ft/${s}^{2}$=3.85 ft/${s}^{2}$ $\rightarrow$